Смирнов В.П. Курс статистической физики. Конспект лекций
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(n, α)
q
n,α
= U
α
exp(i(ka
n
− ωt)),
ω t
α
k U
α
a
i
B
1
= 2π
a
2
× a
3
Ω
a
, B
2
= 2π
a
3
× a
1
Ω
a
, B
3
= 2π
a
1
× a
2
Ω
a
,
B
m
= m
1
B
1
+ m
2
B
2
+ m
3
B
3
.
a
i
i = 1, 2, 3 B
j
j = 1, 2, 3
a
i
· B
j
= 2πδ
ij
.
a
n
· B
m
= 2π
3
X
i=1
n
i
m
i
.
2π
exp(i(a
n
· B
m
)) = 1.
q
n+N
i
a
i
,α
= q
n,α
exp(ik · a
i
) = 1 = exp(i2πm
i
).
k
k = k
1
B
1
+ k
2
B
2
+ k
3
B
3
,
k
i
= m
i
/N
i
k
k
B
i
k
N = N
1
N
2
N
3
k
∆k
i
= N
−1
i
N
i
k 3σ ω
j
(k)
j k ω
j
(k)
ω
j
(k) j
3σ ω
j
(k)
k 3σ
ω
j
(k) → 0 k → 0
ω
j
(q) → ω
(0)
j
k → 0
ω
j
(k) k
r
n,α
= a
n
+ d
α
+ q
n,α
.
q
n,α
3σN q
n,α,i
q
H =
X
n,α,i
p
2
n,α,i
2m
α
+ W (q)
W (q) q
n,α,i
W (q) = W (0) +
X
n,α,i
∂W
∂q
n,α,i
0
q
n,α,i
+
1
2
X
n, α, i;
n
0
, α
0
, i
0
γ
n
0
,α
0
,i
0
n,α,i
q
n,α,i
q
n
0
,α
0
,i
0
+ ···,
γ
n
0
,α
0
,i
0
n,α,i
=
∂
2
W
∂q
n,α,i
∂q
n
0
,α
0
,i
0
0
W (0)
∂W
∂q
n,α,i
0
= 0
H =
X
n,α,i
p
2
n,α,i
2m
α
+
1
2
X
n, α, i;
n
0
, α
0
, i
0
γ
n
0
,α
0
,i
0
n,α,i
q
n,α,i
q
n
0
,α
0
,i
0
, γ
n
0
,α
0
,i
0
n,α,i
= γ
n,α,i
n
0
,α
0
,i
0
.
m
α
¨q
n,α,i
= −
X
n,α,i
γ
n
0
,α
0
,i
0
n,α,i
q
n
0
,α
0
,i
0
, α, α
0
= 1, 2, ..., σ, i, i
0
= x, y, z, n, n
0
∼ N.
q
(k,j)
n,α,i
=
√
m
α
U
(j)
α,i
(k) e xp(i(k · a
n
− ω
j
(k)t)).
ω
j
(k) U
(j)
α,i
X
α
0
,i
0
D
αi,α
0
i
0
(k)U
(j)
α
0
,i
0
(k) = ω
2
j
(k)U
(j)
α,i
(k),
D
αi,α
0
i
0
(k) = (m
α
m
α
0
)
−1/2
X
n
0
γ
n,α,i
n
0
,α
0
,i
0
exp(ik · (a
n
0
− a
n
)).
q
n,α,i
=
X
j,k
Q
j
(k)U
(j)
α,i
(k) e xp(i(k · a)
n
− ω
j
(k)t).
Q
j
(k)
P
j
(k)
H(Q, P ) =
X
j,k
H
j,k
, H
j,k
=
1
2
[P
2
j
(k) + ω
j
(k)Q
2
j
(k)].
(j, k)
kT E = 3σNkT
C
V,N
= 3σNk C
V,N
= 3Nk
C
V,N
= 3N
A
k N
A
= 6.02 × 10
−23
(j, k) ω
j
(k)
n
j,k
E
n
j,k
= ~ω
j
(k)(n
j,k
+ 1/2), n
j,k
= 0, 1, 2, ... .
3σN {n
j,k
}
E
{n
j,k
}
=
X
j,k
E
n
j,k
.
Z =
X
{n
j,k
}
exp
−
E
{n
j,k
}
kT
=
Y
{n
j,k
}
Z
{n
j,k
}
,
Z
{n
j,k
}
=
∞
X
n
j,k
=0
exp
−
E
n
j,k
kT
= exp
−
~ω
j
(k)
2kT
1 − exp
−
~ω
j
(k)
kT
−1
,
E = kT
2
∂ ln(Z)
∂T
=
X
j,k
~ω
j
(k)
2
+
X
j,k
(~ω
j
(k)), (y) =
y
exp
y
kT
− 1
.
E
o
k Ω
B
=
(2π)
3
Ω
a
k N
(2π)
3
V
V
X
k
→
V
(2π)
3
Z
dk.
ω
j
(k)
ω
j
(k) = ω
(o)
j
, j = 4, 5, ...3σ.
T
oj
= ~ω
(o)
j
/k (j = 4, 5, ..., 3σ)
E
opt
=
3σ
X
j=4
X
k
~ω
(o)
j
exp
~ω
(o)
j
/kT
− 1
= NkT
3σ
X
j=4
T
oj
/T
exp(T
oj
/T ) − 1
.
ω
j
(k) = vk, j = 1, 2, 3.
k
k
m
E
ac
=
3
X
j=1
X
k
(~vk) → 3
V
(2π)
3
Z
(~vk)dk = 3
V
(2π)
3
4π
Z
k
m
0
(~vk)k
2
dk.
k → ω ω = vk
E
ac
=
3V
2π
2
v
3
Z
ω
m
0
(~ω)ω
2
dω =
Z
ω
m
0
(~ω)g(ω)dω, g(ω) =
3V
2π
2
v
3
ω
2
.
g(ω)dω dω
3N
3N =
Z
ω
m
0
g(ω)dω =
V ω
3
m
2π
2
v
3
, ω
m
= v
6π
2
Ω
a
1/3
.
E
ac
=
Z
ω
m
0
~ω
exp(~ω/kT ) − 1
g(ω)dω.
x = (~ω/kT )
T
D
= ~ω
m
/k
E
ac
= 3NkT D(T
D
/T ),
D(y)
D(y) =
3
y
3
Z
y
0
x
3
dx
exp(x) − 1
.
E = E
o
+ NkT
3D(T
D
/T ) +
3σ
X
j=4
T
oj
/T
exp(T
oj
/T ) − 1
!
.
T T
D
, T
oj
, T
D
/T = y 1
y
D(y) =
3
y
3
Z
y
0
x
3
dx
exp(x) − 1
≈
3
y
3
Z
y
0
x
3
dx
x + x
2
/2
≈
3
y
3
Z
y
0
(x
2
− x
3
/2)dx = 1 −
3
8
y
T
oj
/T
exp(T
oj
/T ) − 1
≈ 1 −
T
oj
2T
.
E = E
o
+ 3NσkT + O(1/T ).
C
V,N
= 3σNk; σ = 1 C
V,N
= 3Nk,
T T
D
, T
oj
D(y) ≈
3
y
3
Z
∞
0
x
3
dx
exp(x) − 1
=
π
4
15
3
y
3
,
x
exp(x) − 1
≈ x exp(−x) x 1.
E = E
o
+
3π
4
5
NkT (T/T
D
)
3
+ O(exp(−T
oj
/T )),
C
V,N
≈
12
5
π
4
Nk(T/T
D
)
3
∼ T
3
.
ω
j
(k)
T
T
D
T
D
Pb KBr NaCl
T
D
K 90 180 280 2000
ρ
q p
λ t
ρ = ρ(q, p, λ; t) A
A = A(q, p; t)
a
ρ
A
(a, t)
ρ = ρ(q, p, λ; t)
ρ
A
(a, t) =
Z
δ(a − A(q, p; t))ρ(q, p, λ; t)dΓ.
R
ρ
A
(a, t)da =
Z
Z
δ(a − A(q, p; t))da
| {z }
ρ(q, p, λ; t)dΓ =
R
ρ(q, p, λ; t)dΓ = 1.
= 1
ρ = ρ
E
o
(q, p, ; V, N) =
1
g(E
o
, V, N)
δ(E
o
− H(q, p))
E = H(q, p)
ρ
E
o
=
Z
1
g(E
o
, V, N)
δ(E − H(q, p))δ(E
o
− H(q, p))dΓ =
Z
1
g(E
o
, V, N)
δ(E − H)δ(E
o
− H)g(H)dH = δ(E −E
o
).
E =
Z
Eδ(E − E
o
)dE = E
o
, E
2
= E
2
o
, (∆E)
2
= 0.
ρ = ρ
can
(q, p; T, V, N) =
1
Z(T, V, N)
exp(−
H(q, p)
kT
)
E =
H(q, p)
ρ
can
(E) =
Z
1
Z(T, V, N)
δ(E − H(q, p)) exp
−
H(q, p)
kT
dΓ =
=
1
Z(T, V, N)
Z
δ(E − H) exp
−
H
kT
g(H)dH =
1
Z(T, V, N)
exp
−
E
kT
g(E).
Z(T, V, N) =
Z
exp(−E/kT )g(E)dE.
β = (kT )
−1
E =
Z
E exp(−βE)g(E)dE = −
1
Z
∂Z
∂β
, E
2
=
1
Z
∂
2
Z
∂β
2
,
(∆E)
2
=
∂
∂β
1
Z
∂Z
∂β
= −
∂E
∂β
.
β T −
∂
∂β
= kT
2
∂
∂T
E = kT
2
1
Z
∂Z
∂T
= kT
2
∂ ln Z
∂T
, (∆E)
2
= kT
2
C
V,N
, δ
E
=
p
kT
2
C
V,N
E
.
C
V
=
3
2
Nk, E =
3
2
NkT, δ
E
=
r
2
3
N
−1/2
.
δ
E
=
2
√
30.6
π
2
N
−1/2
.
σ
N σN
δ
E
=
1
√
3
(σN)
−1/2
,
δ
E
=
2
√
5σ
√
3π
2
T
D
T
3/2
(σN)
−1/2
.
∆E =
∂E
∂T
V,N
∆T = C
V,N
∆T.
(∆E)
2
= C
2
V,N
(∆T )
2
(∆T )
2
=
(∆E)
2
C
2
V,N
=
kT
2
C
V,N
, δ
T
=
q
(∆T )
2
T
=
s
k
C
V,N
.
ρ
bcan
=
1
Z
b
exp
µ
e
N − E
n,
e
N
kT
!
, Z
b
=
∞
X
e
N=0
exp
µ
e
N
kT
!
X
n
−
E
n,
e
N
kT
.
N =
∞
X
e
N=0
e
N exp
µ
e
N
kT
!
X
n
−
E
n,
e
N
kT
=
kT
Z
b
∂Z
b
∂µ
, N
2
=
(kT )
2
Z
b
∂
2
Z
b
∂µ
2
.
(∆N)
2
= (kT )
2
"
1
Z
b
∂
2
Z
b
∂µ
2
−
1
Z
2
b
∂Z
b
∂µ
2
#
= (kT )
2
∂
∂µ
1
Z
b
∂Z
b
µ
= kT
∂N
∂µ
T,V
.
δ
N
=
s
kT
N
2
∂N
∂µ
T,V
.
∂N
∂µ
T,V
=
∂N
∂p
T,V
∂p
∂µ
T,V
=
N
V
∂N
∂p
T,V
,
∂p
∂µ
T,V
= −
∂
∂µ
∂Ω
∂V
T,µ
= −
∂
∂V
∂Ω
∂µ
T,V
=
∂N
∂V
T,µ
=
N
V
.
(∆N)
2
=
kT N
V
∂N
∂p
T,V
, δ
N
=
s
kT
NV
∂N
∂p
T,V
.
(83) (84)
f(T, V, p, N) = 0
pV = NkT,
∂N
∂p
T,V
=
V
kT
, (∆N)
2
= N, δ
N
= N
−1/2
.
V
2
n =
N
V
(∆n)
2
=
kT N
V
2
∂n
∂p
T,V
=
kT N
V
2
∂n
∂p
T,N
.
∆n = −
N
V
2
∆V,
N
2
V
4
(∆V )
2
= −
kT N
V
2
N
V
2
∂V
∂p
T,N
.
(∆V )
2
= −kT
∂V
∂p
T,N
.
(∆V )
2
=
V
2
N
, δ
V
= N
−1/2
.