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PHY 770 Spring 2014 -- Lecture 18 14/01/2014
PHY 770 -- Statistical Mechanics12:00* - 1:45 PM TR Olin 107
Instructor: Natalie Holzwarth (Olin 300)Course Webpage: http://www.wfu.edu/~natalie/s14phy770
Lecture 18
Chap. 7 – Brownian motion and other non-equilibrium phenomena
Fokker-Planck equation examples Linear response theory Fluctuation-Dissipation Theorem
*Partial make-up lecture -- early start time
PHY 770 Spring 2014 -- Lecture 18 24/01/2014
PHY 770 Spring 2014 -- Lecture 18 34/01/2014
PHY 770 Spring 2014 -- Lecture 18 44/01/2014
PHY 770 Spring 2014 -- Lecture 18 54/01/2014
Probability analysis of Brownian motion -- Fokker-Planck equationMacroscopic Microscopic
probability: probability:
( , , ) ( , , , )P x v t x v t
2
2 2
Fokker-Planck equation
1(
2
:
)P P g P
v v F x Pt x v m m m v
1 22 1
Stochastic for
( ) )
e
(
c
) (t t g t t
Conservative force
( )( )
dV xF x
dx
friction coefficient
PHY 770 Spring 2014 -- Lecture 18 64/01/2014
( ) 1 1 ( )( ) ( ) ( ) ( )
Corresponding Langevin equation
:
dv t dx tv t F x t v t
dt m m m dt
PHY 770 Spring 2014 -- Lecture 18 74/01/2014
Example solution in the limit of large friction
( ) ( )
( ) 1 1 ( )( ) ( ) ( ) ( )
When is sufficiently large, the system reaches a steady-sta
Langevin equation in presence of friction ( ) and
potential for
te
ver
ce ( )
y r
:F x x
dv t dx tv t F x t v t
dt m m dt
V
m
2
2 2
( )apidly so the 0 Then the Langevin equation
( ) 1 1reduces to: ( ) ( )
The Fokker-Planck equation reduces to:
1( )
2
dv t
dtdx t
F x tdt
P g PF x P
t x x
PHY 770 Spring 2014 -- Lecture 18 84/01/2014
Example solution in the limit of large friction -- continued
2
2 2
2 2
2 2 2 2 2
In this case:
1( )
2
Further consider the case where ( ) 0 :
2 Note:
2 2 2
P g PF x P
t x x
F x
P g P P g kT kTD D
t x x
2
2
2
2
4
Solution of this diffusion equation:
1Let ( , ) ( , )
2
( , )( , ) ( , )
( , ) 2 4
iqx
Dq t
xiqx Dq t Dt
P x t dqe P q t
P q tq DP q t P q t Ce
t
C CP x t dqe e e
Dt
PHY 770 Spring 2014 -- Lecture 18 94/01/2014
Example solution for “free” particle V(x)=0
2
2 2
2
2 2
F
1 ( )
2
For ( ) 0, we can assume ( , ,
okker-Planck equation
)
:
( , )
2
P P dV x g Pv v P
t x v m m dx m v
V x P x v t P v t
vPP g P
t m v m v
2 (
2 2
2
/ 2 ) ( , )In order to find solution, define ( , )
1Let
( , )(:
2 2, )
4
m v g v t
v tv t
m t
P v t e
gA A
m A v
v
PHY 770 Spring 2014 -- Lecture 18 104/01/2014
Example solution for “free” particle V(x)=0
2
2
2 2 2
2 2
2
/(4 )
/
0
/(4 ) /
0
/(2 ) / /(
( ) ( )
1where ( )
2
( , ) ( )
1Note t
( , )
hat: 2 4
( )
For ( , )
2! 2
v
v
v
n n
An n
n
nt mn n
n
A nt mn n
n
A m v g mv
v n vv
v
vA
A
ve
An
a e
e a e
HA
v t v
P v t v
t P v t K Ke e Ke
2 )
when 2
kT
g kT
PHY 770 Spring 2014 -- Lecture 18 114/01/2014
Fluctuations about equilibrium
0
0
0 0
Consider a Brownian particle of mass m in the presence of
random noise , fluid friction , and ( ) ( ) :
( )+ ( ) ( )
( ) ( )
an an imp
( ) ( ) ( )
Solution:
ulse t t
dvm v t t t
dtd v t d t
m v t t v tdt d
F
F
Ft
v
F
/00( ) ( )e ( ) t mt t K t
FF
m
response function/e
1( ) ( )t mK t t
m
PHY 770 Spring 2014 -- Lecture 18 124/01/2014
Linear response function
We can define the linear response ( ) of a variable ( ) to an
external force (
( ') for '' ( ') ( ') where ( ')
0 fo '
)
)
(r
:
F
K t t t tdt K t t F t
K t t
F
K t tt t
t
t
1 1e ( ) ( )e
2 2
( ) ( )e ( ) (
Four
)e
ier transforms
( ) ( )
( )
2
( ) ( )
i t i t
F F
i t i t
F
d F t d F
K t d d K t
t
t
F
PHY 770 Spring 2014 -- Lecture 18 134/01/2014
Linear response function -- simple example /e1
( ) ( )t mK t tm
/ ( )
1 = for this example.
Note: The pole of
More generally:
1 ( ) ( )e e
( ) has 0.
( ) ( ) ( )
The real and imaginar
y
parts of ( )
i t t m i t
R I
d K t dt et
z z
i
i
t
m
m
satisfy the Kramers-Kronig relat
u
ions:
( )1( )
( )
u 1
( )
IR
RI
uP d
u
u
uP d
PHY 770 Spring 2014 -- Lecture 18 144/01/2014
“Proof” of Kramers-Kronig relations
z-α
f(z)dz
πifzfdz
zf
includes
2
1 0)(
:)(function analytican for formula integral sCauchy'Consider
Re(z)
Im(z)
a
PHY 770 Spring 2014 -- Lecture 18 154/01/2014
Kramers-Kronig transform -- continued
z-α
f(z)dz
-αz
)f(zdz
πi
z-α
f(z)dz
πif
restR
RR
includes
2
1
2
1
Re(z)
Im(z)
a
=0
f-αz
)f(zdzP
πi
-αz
)f(zdz
πi f
R
RR
R
RR )(
2
1
2
1
2
1
PHY 770 Spring 2014 -- Lecture 18 164/01/2014
Kramers-Kronig transform -- continued
-αz
)(zfdzPf
-αz
)(zfdzPf
-αz
zifzfdzP
πiiff
zifz fz f
f-αz
)f(zdzP
πi f
R
RRRI
R
RIRR
R
RIRRRIR
RIRRR
R
RR
1
1
2
1
2
1
: Suppose
)(2
1
2
1
PHY 770 Spring 2014 -- Lecture 18 174/01/2014
Kramers-Kronig transform -- continued
-αz
)(zfdzPf
-αz
)(zfdzPf
R
RRRI
R
RIRR
1
1
This Kramers-Kronig transform is useful for the susceptibility function
when
Must show that: 1. is analytic for 0
2. van
( )
ishes for
R
I
f z
f z z
f z z
1For the example: = requirements are me) t( .
im
PHY 770 Spring 2014 -- Lecture 18 184/01/2014
Fluctuation-Dissipation TheoremRelationship of the response and correlation functions of a system near equilibrium; allows a weak external field to probe equilibrium fluctuations
0 0
0
0 0
0 0
0 0 0
Consider a function ( ). A force is applied:
0 0( )
0
Let ( , ) represent the probabil
in the presence of .
Generally, we expe
ity distribution
for (0)
( ) ( , ) (
c :
)
t
F
t
tF t
F t
P F
d
F
t P F t
0
0 for 0( ) Mtt e t
PHY 770 Spring 2014 -- Lecture 18 194/01/2014
Fluctuation-Dissipation Theorem -- continued
0 0
0 0 0
0 0 0
0 for 0
Relationship to response function:
1' ( ') ( ')
2
( ) ( , ) ( )
( ) ( ) (0)
( ) ( )
F
Mt Mt
F
i t
F F
d
e t e
dt K t t F t
t P F t
d
t t
t e
( (1
2) = )i ted F
0
0
0 0For: ( )
0
(1
)
tF t
F t
F F Pi
HW #17
PHY 770 Spring 2014 -- Lecture 18 204/01/2014
Fluctuation-Dissipation Theorem -- continued
0
0
0
( )
( ) ( ) ( )
1
( )cos
( )
(0)
1=
2
0
0
i t
F
F
F F
t e
Pi
d
F
F
t
tP d t
i
tF
HW #17
PHY 770 Spring 2014 -- Lecture 18 214/01/2014
Fluctuation-Dissipation Theorem -- continued
0 0
0 0
0
0 0 0
0
0
( ) ( , ) ( )
( ) (0)
( ) cos
( )
(0)
for 0
0
0
F
Mt
F
F
d
e t
t P
FP d
F t
t
t
tt
i
F t
( )(0)
1Note that: P d
i
1 ( )cos
(0)Mte P di
t
PHY 770 Spring 2014 -- Lecture 18 22
21
2
0 0
Note that: (0)
( ) cos( ) (0)
(
1
0
1
)P d
i
ST
tt
4/01/2014
Fluctuation-Dissipation Theorem -- continued
0 0
0 0
0 0
0 0
Correlation function:
1
( ) (0)
( ) (0) (0) (0)
( ) cos
( ) cos( ) (0)
(0
=
1
)
1
Mt
Mt
e
e
P di
P
t
t
t
td
it
PHY 770 Spring 2014 -- Lecture 18 234/01/2014
Fluctuation-Dissipation Theorem -- example
0 0
( ) cos( ) (0)
(0)
1 1 tdt P
i
0
0
0 0
Consider a Browian particle of mass m in the presence of
random noise , fluid friction , and a( ) ( ) :
( )+ ( ) ( )
( ) ( )( ) ( ) ( )
Solution:
n an impul
(
se t t
dvm v t t t
dtd v t d t
m v t t v tdt dt
F
v
F
F
F
/00) ( )e ( ) t mt t
FF
mK t
/ ( )
1 = for this example.
1 ( ) ( )e e i t t m i td K t d e
mt t t
im
PHY 770 Spring 2014 -- Lecture 18 244/01/2014
Fluctuation-Dissipation Theorem -- example
1 =( )
im
0 0
( ) cos( ) (0)
(0)
1 1 tdt P
i
20 0 0
/
( ) cos1
( ) (0
) ( ) (
=
)
0
t m
kTv
m
kTP d
m i
kTe
tt v t
m
v
PHY 770 Spring 2014 -- Lecture 18 254/01/2014
Microscopic linear response theory
0
0
Consider a Hamiltonian consisting of representing the system
in absence of a field and a small pert :
( ) ( ) ( )
In terms of the d
urbing field described b
ensity opera
y
tor:
( )( ), ( )
As
H
H t H t H t
ti t
H
H tt
sume: ( ) ( ) ( ) where ( )
Tr
o
o
H
eq eq H
et t t t
e
0
( )( ), ( ) ( ), ( ) ( ), ( )eq
ti H t t H t t H t t
t
0
PHY 770 Spring 2014 -- Lecture 18 264/01/2014
Microscopic linear response theory -- continued
0 0
0
( )/ ( )/
( )( ), ( ) ( ), ( )
Assume ( ) 0
1( ) ' ( ) , ( )
eq
tiH t t iH t t
eq
ti H t t H t t
tt
t dt e H t e ti