Di usion et marche aléatoire

θ±∆L
−→
Ux+−1
2
1
2
hx0i=1
2(+∆L) + 1 2(−∆L) = 0
hxNi=Nhx0i= 0
x2
0=1
2(+∆L)2+1 2(−∆L)2= ∆L26=hx0i2
x2
N=Nx2
0=N∆L2
T N =T
θ
phx2
Ni=qT
θ∆L
d'√DT
x t
Pm,N xm=m∆L N
Pm,N =nombre de possibilites d0obtenir xm=m∆L en N sauts
nombre de possibilites totales en N sauts

N2N
xmp N −p
p∆L−(N−p)∆L=m∆L p =N+m
2
N
pxmN
Pm,N =
N
p
2N=1 2N
N!
p!(N−p)! =1
2N
N!
(N+m
2)!(N−m 2)!
N1m1
ln (Pm,N ) = ln 1
2N+ln (N!) + ln 1
(N+p
2)!+ln 1
(N−p
2)!
=−Nln (2) + N lnN −N+1
2ln (2πN)−N+m 2ln N+m 2+N+m 2−1 2ln (π(N+m)) −N−m 2ln N−m 2+
N−m
2−1 2ln (π(N−m))
=−Nln (2) + N lnN +1
2ln (2πN)−N+m 2ln N+m 2−1 2ln (π(N+m)) −N−m 2ln N−m 2−1 2ln (π(N−m))
=−Nln (2) + N lnN +1
2ln (2πN)−N+m 2ln N 21 + m N−1 2ln πN 1 + m N−N−m 2ln N 21−m N−
1
2ln πN 1−m N
=−Nln (2) + N lnN +1
2ln (2πN)−N+m 2lnN −ln2 + ln 1 + m N−1 2lnπN +ln 1 + m N−
N−m
2lnN −ln2 + ln 1−m N−1 2lnπN +ln 1−m N
=1
2ln (2πN)−N+m 2ln 1 + m N−1 2lnπN +ln 1 + m N−N−m 2ln 1−m N−1 2lnπN +ln 1−m N
=1
2(ln (2πN)−2lnπN)−N+m+1 2ln 1 + m N−N−m+1 2ln 1−m N
=1
2ln 2 πN −N+m+1 2m N−m2 2N2−N−m+1 2−m N−m2 2N2
ln (Pm,N ) = 1
2ln 2 πN −m2 2N
Pm,N =q2
πN exp −m2 2Nln (N!) 'N lnN −N+1
2ln (2πN )

x x+dx t
P(x, t)dx
P(x, t)dx =Pm,N +Pm+1,N +... +Pm+k,N m=x
∆Lk=dx ∆LN=t θ
Pm,N Pm,N Pm+k,N
P(x, t)dx =kPm,N =Pm,N dx
∆L
P(x, t) = q2
πN∆L2exp −m2 2NN m
P(x, t) = q2θ
πt∆L2exp −x2θ 2t∆L2=1
√4πDt exp −x2
4Dt 
D=∆L2
2θ
x= 0
(∂n
∂t −D∂2n ∂x2= 0
n(t= 0, x) = δ(x)
hX(t)i¯ X(t)
−→
f=−αM−→ v
−→
F(t)
D−→
F(t)E= 0
hFi(t)Fj(t+τ)i= 2D δij δ(τ)
Md−→
v
dt =−αM−→
v+−→ F(t)
−→
v(t) = −→ v(0)e−αt +1
M´t
0−→
F(u)eα(u−t)du
t
h−→ v(t)i=−→ v(0)e−αt+1
Mˆt
0D−→
F(u)eα(u−t)Edu
=−→ v(0)e−αt +1
Mˆt
0D−→
F(u)Eeα(u−t)du
=−→ v(0)e−αt

t
1
2M−→
v(t)2=*1
2M−→
v(0)e−αt +1
Mˆt
0
−→
F(u)eα(u−t)du2+
=1
2M−→
v(0)2e−2αt+1
2M*ˆt
0
−→
F(u)eα(u−t)du2++−→ v(0)e−αt ˆt
0
−→
F(u)eα(u−t)du
=1
2M−→
v(0)2e−2αt +1
2Mˆt
0
du ˆt
0
dv D−→ F(u)−→ F(v)Eeα(u+v−2t)+−→ v(0)e−αt ˆt
0D−→
F(u)Eeα(u−t)du
=1
2M−→
v(0)2e−2αt +1
2Mˆt
0
du ˆt
0
dv 2Dδ(v−u)eα(u+v−2t)
=1
2M−→
v(0)2e−2αt +D
Mˆt
0
du e2α(u−t)
=1
2M−→
v(0)2e−2αt +D
2Mα 1−e−2αt)
hEci ' D
2Mα
hEci ' 3×kBT
2⇒D= 3MαkBT