Смирнов В.П. Курс статистической физики. Конспект лекций
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J
j
= V
1 − (j − 1)
ω
V
, j = 1, 2, ..., N.
Q
N
=
N
Y
j=1
1 − (j − 1)
ω
V
, Z = Z
N
Y
j=1
1 − (j − 1)
ω
V
.
F = −kT ln Z = F − kT ln Q
N
.
Nω
V
1
ln Q
N
=
N
X
j=1
ln
1 − (j − 1)
ω
V
≈ −
N
X
j=1
(j − 1)
ω
V
= −
ω
V
N(N − 1)
2
.
F = F +
N(N − 1)kT
2V
ω(T )
E = E −
N(N − 1)kT
2
2V
∂ω(T )
∂T
=
3
2
NkT
1 −
n − 1
3V
T
∂ω(T )
∂T
.
p = −
∂F
∂V
= p +
N(N − 1)kT
2V
2
ω(T ) =
NkT
V
1 +
(N − 1)ω
2V
.
d
U(r) =
∞ r < d
U
r>d
(r) kT r > d
ω(T ) = 4π
Z
d
0
r
2
dr + 4π
Z
∞
d
(1 − exp
−
u(r)
kT
)r
2
dr = 8v +
V
kT
w,
v
w =
4π
V
Z
∞
d
U(r)r
2
dr
ω
p =
NkT
V
1 +
4v(N − 1)
V
+
N − 1
kT
w
p +
a
V
2
(V − b) = NkT,
p 1/V
p =
NkT
V
1 +
b −
a
NkT
V
+
b
2
V
2
+ ···
.
p =
NkT
V
1 +
B(T )
V
+
C(T )
V
2
+ ···
.
B(T ) C(T )
B(T ) = b−
a
NkT
C(T ) = b
2
b = 4(N − 1)v ≈ 4Nv,
a
V
= −
N(N − 1)
2
w.
b a/V
N
N(N − 1)
2
∂ω
∂T
= −
V
kT
2
w, E =
3
2
NkT +
N(N − 1)
2
w (w < 0),
∆q∆p ≥
~
2
~ =
h
2π
h ~ = 1.05×10
−34
·
m
E = H =
p
2
2m
+
mω
2
2
q
2
.
E ≥ 0
q(t) p(t)
E
n
= ~ω(n + 1/2), n = 0, 1, 2, ...
m
r I = mr
2
ϕ
E = H =
I ˙ϕ
2
2
=
p
2
ϕ
2I
, p
ϕ
= I ˙ϕ.
p
ϕ
ϕ
m
m = 0, ±1, ±2, ...
E
m
=
~
2
m
2
2I
, m = 0, ±1, ±2, ...
m = 0
m r
I = mr
2
θ
ϕ
E = H =
p
2
θ
2I
+
p
2
ϕ
2I sin
2
θ
, p
θ
= I
˙
θ, p
ϕ
= I sin
2
θ ˙ϕ.
E
l
=
~
2
l(l + 1)
2I
, l = 0, 1, 2, ...
2l + 1
Z M
z
M
z
= ~m, m = 0, ±1, ±2, ..., l.
h
3N
3N
N! h
3N
N
h
3N
N! dqdp
dΓ =
dqdp
h
3N
N!
1
h
3N
N!
ρ(E
α
) α ≡ (i, n)
i n
X
α
ρ(E
α
) =
X
i,n
ρ(E
i
) =
X
i
g
i
ρ(E
i
) = 1,
g
i
E
i
Z
ρ(E)g(E)dE = 1,
g(E)dE dE
ρ
qmc
(α, E, V, N) =
1
g(E, V, N)
δ
E,E
α
, ρ
qmc
(α, E, V, N) =
1
g(E, V, N)
δ(E − E
α
)
∆W = g(E, V, N)∆E
S(E, V, N) = k ln g(E, V, N)∆E.
ρ
qc
(α, T, V, N) =
1
Z
exp
−
E
α
(V, N)
kT
.
Z(T, V, N) =
X
α
exp
−
E
α
(V, N)
kT
=
X
i
g(E
i
, V, N) exp
−
E
i
(V, N)
kT
.
Z
ρ
qgc
(α,
e
N, T, V, µ) =
1
Z
qgc
exp
µ
e
N − E
α
(V,
e
N)
kT
!
.
Z
qgc
= Z
qgc
(T, V, µ)
Z
qgc
(T, V, µ) =
∞
X
e
N=0
e
µ
e
N
kT
X
α
e
−
E
α
(V,
e
N)
kT
=
∞
X
e
N=0
e
µ
e
N
kT
X
i
g(E
i
, V,
e
N)e
−
E
i
(V,
e
N)
kT
.
H =
N
X
j=1
H(q
j
, p
j
); H(q
j
, p
j
) =
p
2
j
2m
+
mω
2
2
q
2
j
.
Z = Z
N
o
; Z
o
=
e
Nh
3
Z
∞
−∞
exp
−
p
2
j
2mkT
!
dp
j
Z
∞
−∞
exp
−
mω
2
2kT
q
2
j
dq
j
=
=
e
Nh
3
(2πmkT )
1/2
2πkT
mω
2
1/2
=
2πe
Nh
3
kT
ω
.
E = kT
2
∂ ln Z
∂T
= NkT
C =
∂E
∂T
= Nk.
Z
o
T
1/2
E = E
c
+ E
pot
=
1
2
NkT +
1
2
NkT ; C = C
c
+ C
pot
=
1
2
Nk +
1
2
Nk.
α n
j
α ≡ (n
1
, n
2
, ..., n
N
) ≡ [n
j
]
E
[n
j
]
=
P
N
j=1
~ω(n
j
+ 1/2)
n
j
N!
n
1
, n
2
, ..., n
N
N! ≈
N
e
N
Z =
e
N
N
X
[n
j
]
exp
E
[n
j
]
kT
=
N
Y
j=1
∞
X
n
j
=1
e
N
exp
−
~ω
kT
(n
j
+
1
2
)
= Z
N
o
;
Z
o
=
e
N
∞
X
n
j
=0
exp
−
~ω
kT
(n
j
+
1
2
)
=
e
N
exp
−
~ω
2kT
1 − exp
−
~ω
kT
=
=
e
N
exp
−
T
o
2T
1 − exp
−
T
o
T
=
e
N
exp
−
1
2τ
1 − exp
−
1
τ
,
T
o
=
~ω
k
τ =
T
T
o
E = N~ω
1
2
+
1
exp
T
o
T
− 1
= N~ω
1
2
+
1
exp
1
τ
− 1
C = Nk
T
o
T
2
exp
T
o
T
exp
T
o
T
− 1
2
= Nk
1
τ
2
exp
1
τ
exp
1
τ
− 1
2
τ
C τ
T T
o
τ 1
E = N~ω
1
2
+ exp
−
T
o
T
, C = Nk
T
o
T
2
exp
−
T
o
T
;
T T
o
τ 1
E = NkT
o
1
2
+
T
T
o
≈ NkT, C = Nk.
T
o
ω ∼ 10
13
÷10
14
T
o
∼ 10
3
K
T
o
T
o
N
2
O
2
H
2
HCl HBr
T
o
K 3340 2230 5800 4140 3700
3
H =
N
X
j=1
p
2
ϕ
j
2I
.
Z = Z
N
o
; Z
o
=
e
hN
Z
2π
0
dϕ
Z
∞
−∞
e
−
p
2
ϕ
2IkT
dp
ϕ
=
2πe
hN
(2πIkT )
1/2
,
E = kT
2
∂ ln Z
∂T
=
1
2
NkT,
C =
1
2
Nk.
1
2
kT
1
2
k I
α ≡ (m
1
, m
2
, ..., m
N
) ≡ {m
j
}
E
{m
j
}
=
N
X
j=1
~
2
m
2
j
2I
.
Z = Z
N
o
; Z
o
=
e
N
∞
X
m=−∞
e
−
~
2
m
2
2IkT
=
e
N
∞
X
m=−∞
e
−
T
r
T
m
2
=
e
N
∞
X
m=−∞
e
−
1
τ
m
2
.
T
r
≡
~
2
2Ik
τ ≡
T
T
r
C = kN
∂
∂τ
τ
2
∂ ln Z
o
(τ)
∂τ
.
τ
C τ