PHY 4523
“Statistical Physics” Spring 2003
Formula and Fact Sheet (Version of 07 April 2014)
Professor Mark W. Meisel
Page 1
kB = 1.381 × 10−23 J/K = 8.617 × 10−5 eV/K
[( )
12
−
d
r
V (r) = 4 ϵ
( )6 ]
3
−u
x2
2!
ln(1 + x) = x −
+
x3
3!
+ ... =
x2
2
+
x3
3
−
x =
∫∞
x4
4
∑
i
(
)
pi = limN →∞ nNi
!
W = n1 !×n2N!×n
3 !×...
pi = 1
n!
n
Cr = r!(n−r)!
i
(∆x)2 =
pi x i
e−(x−x)
∑
i
PL (x) =
1
2
(2π) ∆x
x
PP (x) =
x −x
e
x!
PB (x) =
n!
px (1 − p)n−x
x!(n − x)!
Z = j e−E(j /kB T )
ln Z
U = kB T 2 ∂ ∂T
pi =
β =
∑
P = −
(
)
∂F
( ∂V ) T
S = − ∂F
∑∂T V
d′ Q = i dpi Ei
)1/2
(
h2
λD = m2π¯
kB T
V
(
)
∂(T ln Z)
( ∂V ) T
kB ∂(T∂Tln Z)
V
P = kB
S =
∑
U = i pi Ei
√
π
pi,j = pi pj
e−Ei /kB T
Z
1
kB T
(
µB
kB T
)
V
F = −kB T ln Z
dF = −SdT − P dV
S = UT + kB ln(Z)
∑
∑
dU = i dpi Ei + i pi dEi
Z = Ztrans × Zvib × Zrot
∫ ∞
1∫∞
−E(px ,x)/kB T
Z =
dx
dpx e
h −∞
−∞
∫
Z =
0
∞
(2n)!
n!
1
Γ/2
π (x − x)2 + (Γ/2)2
M = µ(n+V−n− ) = µN
tanh
( V)
∑
∂Ei
Mz = − V1 i pi ∂B
z
Z = n gn e−En /kB T
ln Z
U = − ∂ ∂β
dx =
p(i+j) = pi + pj
S = ϕ(W )
S = kB ln(W )
∆UA2 = − kSB′′ = kB T 2 CA
∑
π
2
pi (xi − x)2
2 /[2∆x2 ]
PG (x) =
2
√
vrms = v 2
P V = 31 N m v 2
∫
∑
|u| < ∞
|u| < ∞
1 ∫ ∞ 3 ∫ ∞ 3 −E(⃗p,⃗r)/kB T
d ⃗r
d p⃗ e
Z = 3
h −∞
−∞
e−ϵ(k)/kB T D(k) dk
π
2
√
e−x dx =
0
∫ ∞ 2n −(x/a)2
0 x e
|x| < 1 , x = 1
+ ... ,
v 2 = 0∞ v 2 f (v) dv
P = 13 n m v 2
ϵ = f2 kB T
x =
2n+1
xn
n=0 n!
∫
n!
∑ (n−r)!
n x
i i i
N
5
2 5
17 7
62
tanh u = u − 13 u3 + 15
u − 315
u + 2835
u9 − ... |u| <
d
d
(cosh u) = du
(sinh u)
(tanh u) =√du
(sech2 u)
dx
dx
dx
dx
∫
∑∞
∞
−nx
−ax2
= 1−e1−x
dx = πa
n=0 e
−∞ e
∑∞
v = 0∞ v f (v) dv
Φ = 14 n v
3
k T = 12 mv 2
2 B
pi = W1
n
Pr =
R
NA
u
sinh u = u + u3! + u5! + ... + (2n+1)!
+ ...
2
4
2n
u
cosh u = 1 + u2! + u4! + ... + (2n)!
+ ...
−e )
tanh u = (e
(eu +e−u )
d
(sinh u) = du (cosh u)
∫dx∞ 2n+1 −axdx
2
e
dx = 2an!
n+1
0 x
ex = 1 + x +
kB =
ln(M !) ≈ M ln(M ) − M
d
r
sinh u = 12 (eu − e−u )
cosh u = 12 (eu + e−u )
u
1 eV = 1.602 × 10−19 J
( )2n+1
a
s
PHY 4523
“Statistical Physics” Spring 2003
Formula and Fact Sheet (Version of 25 February 2004)
Professor Mark W. Meisel
ZN ≃
Z1N
N!
∫∞
f (k) A(k) dk
0∫
∞
A =
e
D(k) dk =
ω = ck
−1
dQ
dt
f (k) dk
Ak
dk
2π
0
2
D(k) dk = V2πk2 dk
n(⃗k) = h¯ ω(⃗k)/k1 B T
Page 2
D(k) dk = Lπ dk
2
CV = −T ( ∂∂TF2 )V
∫
F = −kB T ⃗k ln Z(⃗k) = kB T 0∞ D(k) ln(1 − e−¯hω/kB T )dk
∑
A ∫∞
U = ⃗k h
U = 2π
¯ ω(k)n(k)kdk
¯ ω(k)n(⃗k)
0 h
∑
−¯
hωα /kB T
F = N ϵ + kB T α ln(1 − e
)
∑
−¯
hωE /kB T
F = N ϵ + 3N kB T ln(1 − e
)
(
∂S
∂N
)
U,V
µchem = −T
(
dU = T dS
( − P dV
) + µdN
∂ ln(WR )
1
= kB
T
∂UR
µ = ∆ + kB T
Ξ =
∑
j
e
NR
ln( nn ′ )
Q
(
CV = 3N kB
D(ω)dω = 2πV 2 ω 2 ( s13 + s23 )dω
T
∫L
−¯
hω/kB T
F = N ϵ + 3k2πB2Ts3V 0∞ ω 2 ln(1 − e
)dω
3
∫
V
ω
ω
3N = 2π3V2 s3 0 D ω 2 dω = 2π2 Ds3
∫
−¯
hω/kB T
F = N ϵ + 3k2πB2Ts3V 0ωD ω 2 ln(1 − e
)dω
(
)3 ∫
4
x
XD x e
CV = 9N kB ΘTD
dx
0
(ex −1)2
µ = −T
)
∂S
∂n U,V
3
s3
=
( s13
L
+
2
s3T
)2
hωE
¯
kB T
XD =
µ =
(
∂U
∂N
−(Ej −µNj )/kB T
(e
pi =
Ξ
ΦG = U − µN − T S = −kB T ln(Ξ)
(
∂ΦG
S = −
∂T
(
∂ΦG
P = −
∂V
(
N = −
∂ΦG
∂µ
4 V
2π 2 kB
5¯
h5 s3
(
3kB T V
2π 2 s3
kB T
h
¯
)3 ∫
ωD /T
0
)
µ =
S,V
(
)
∂G
∂N T,P
(
=
V,µ
)
(
=
T,µ
)
1
λ3D
=
(
mkB T
2π¯
h2
1
+1
=
T,V
1
e[ϵ(k)−µ]/kB T
+1
∂(kB T ln(Ξ))
∂T
∂(kB T ln(Ξ))
∂V
(
−x
x2 ln(1 − e )dx
µ = µint + ∆(⃗r)
W = i Wi
dΦG = −SdT − P dV − N dµ
)
T3
ΘD
T
(ϵi −µ)/kB T
∑
−(Ei −µNi )/kB T
−1)2
)
nQ =
e
−1)
h
¯ ωE /kB T
dG = −SdT( + V dP
) + µdN
∂ ln(WR )
µ = −kB T
∂NR
ni =
hc/λkB T
e
UR
e
λ5 (e
h
¯ ωE /kBT
ωD = s(6π 2 N
)1/3
V
F = Nϵ +
8πhc
u(λ) =
CV =
nQ′ = gnQ
nF D = n(k) =
σ = 5.67 × 10−8 W m−2 K−4
= Aσ T4
)
V,µ
)
∂(kB T ln(Ξ))
∂µ
T,µ
)
nBE = f (k) =
T,V
1
e[ϵ(k)−µ]/kB T
−1
)3/2
PHY 4523
“Statistical Physics” Spring 2003
Formula and Fact Sheet (Version of 24 April 2003)
Professor Mark W. Meisel
nF D = n(k) =
N = 2
∑
1
e[ϵ(k)−µ]/kB T
⃗
⃗k n(k)
2 2
h k
¯
2m
2
1/3
(
)3/2
2mϵ
V
2
2
3π
h
¯
2
CVel = 31 π 2 D(ϵF ) kB
N =
+1
2
kB Tc = 3.31 ¯hm
(N − N0 ) =
I = −
W
−1
∑
e[ϵ(k)−µ]/kB T
nM B = e−[ϵ(k)−µ]/kB T
−1
∫
V k2
1
k 2 dk
n = N
= π12 0∞ [ϵ(k)−µ]/k
B T +1
2π 2
V
e
∫
N = 2 2πV 2 0kF k 2 dk = 3πV 2 kF3
kF
h2
¯
vF = ¯hm
EF = ϵF = 2m
(3π 2 n)2/3 = kB TF
( )3/2
= 2πV 2 2m
D(ϵ) = dN
ϵ1/2
dϵ
( ¯h2 )
2
1 2
π N kB TTF = N k2B π kEBFT
2
T =
µ = kB T ln(nλ3D )
¯ 2 k2
h
2m
1
nBE = f (k) =
D(k) =
ϵ(k) =
kF = (3π n)
ϵ(k) =
Page 3
(
D(k) =
2π 2 N
V
)2/3
N
V
1
V ∫∞
2π 2 0 e[ϵ(k)−µ]/kB T −1
=
(
i=0
N =
)
2mkB Tc 3/2
2
h
¯
k 2 dk = N
S = −kB
pi log2 pi
V k2
2π 2
∑
(
T
Tc
∫∞
0
V
2π 2
∑
A(n) Pe (n)
n
∂t P (n, t) =
∑
n′
0
1
e
[ϵ(k)−µ]/kB T
−1
k 2 dk
x2
2
ex −1
dx
κ =
kB
J 1
= 9.57 × 10−24
log2 e
K bit
)3/2
pi ln pi
i
⟨A⟩e =
∫∞
⟨A(t)⟩ =
∑
A(n) P (n, t)
n
[ Wn′ n P (n′ , t) − Wnn′ P (n, t) ]

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Boltzmann統計力学