PHY 770 -- Statistical Mechanics 12:00 * - 1:45 P M TR Olin 107

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

http://www.wfu.edu/~natalie/s14phy770





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


( ) 1 1 ( )( ) ( ) ( ) ( )

Corresponding Langevin equation

:

dv t dx tv t F x t v t

dt m m m dt


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


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


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


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


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


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


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


“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


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



-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



-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


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


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



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



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


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


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


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


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


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

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PHY 770 -- Statistical Mechanics 12:00 * - 1:45 P M TR Olin 107