Distribution de Boltzmann : échantillonnage

N
T E
µ
pµ=Ce−
Eµ
kT
Eµk C
1
X
µ
pµ= 1
pµ=e−
Eµ
kT
Pµ0e−
Eµ0
kT
N
pµ
pµ0
=e−
Eµ−Eµ0
kT
kT
kT

E N > 1
N
Ω(N, E) = EN
E
p(E) = CΩ(E)e−E
kT
Ω(E)E
Ω(E) = 1
Xµ
X=PµXµe−
Eµ
kT
Pµe−
Eµ
kT
E=PµEµe−
Eµ
kT
Pµe−
Eµ
kT
X
var(X) = ((X−X)2)
var(X) = X2+X2−2XX =X2−X2
X
∆X=pvar(X)
∆E
E

β=1
kT
Z=X
µ
e−βEµ
Z Z
E=−∂ln Z
∂β N,V
∆E2=∂2ln Z
∂β2N,V
=−∂E
∂β N,V
=kT 2Cv
∆E
E=√kT 2Cv E
N
∆E
E'O1
√N
τ
τ

N
M0M−1m
m
m
m= (m+1
2)~ω
Etat Energie
0
1
2
3
4
5
ε1
ε2
ε3
ε4
ε5
ε0
Particule 1 Particule 2 Particule 3 Particule 4 Particule 5 Particule N
..........
ε
m1=1 m2=0 m3=3 m4=1 m5=2

m
pm=e−
m
kT
PM−1
m0=0 e−
m0
kT
e=PM−1
m=0 me−
m
kT
PM−1
m=0 e−
m
kT
z=
M−1
X
m=0
e−β(0+m)=e−β0
M−1
X
m=0 e−βm=e−β01−e−βM 
1−e−β
e=0+
eβ −1−M eβM −1
kT
M→ ∞
∆e2=2eβ
(eβ −1)2−(M)2eβM (eβM −1)2
k= 1
= 1
M