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

PHY 770 Spring 2014 -- Lecture 16 13/25/2014

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 16

Chap. 7 – Brownian motion and other non-equilibrium phenomena

Overview Langevin equation Correlation function and spectral density Fokker-Planck equation

*Partial make-up lecture -- early start time

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





http://famousbiologists.org/robert-brown/

http://famousbiologists.org/robert-brown/


Brownian motion:Phonomenon: Under a microscope a large particle (~1 m in diameter) immersed in a fluid with the same density as the particle, appears to be in a state of agitation, undergoing rapid and random motions.

http://upload.wikimedia.org/wikipedia/commons/c/c2/Brownian_motion_large.gif

http://upload.wikimedia.org/wikipedia/commons/c/c2/Brownian_motion_large.gif

73/25/2014 PHY 770 Spring 2014 -- Lecture 16

Brownian motion:Description based on the Langevin equation of motion

Consider a particle of mass and radius immersed in a fluidof particles (of mass much smaller than ) undergoing Brownianmotion. The fluid gives rise to a retarding force (friction) that isproport

m am

ional to the velocity ( ) and a random force due to the random density fluctuations in

( )

( )

the fluid.

1( ) ( )

t

dv t v tdt m

v

t

t

m

friction coefficient


Brownian motion and Langevin equation of motion -- continued

1 22 1

( ) 1( ) ( )

Properties of random force:( )

( ) ( )

0

( )

dv t v t tdt m m

t

t gt t t

0 0

( / ) ( / )( )0

0

( / ) ( / )( )0 0

0

;

1 (

For initial conditions

)

1 ( )

:( ) ( )

( )

( ) 1 1

tm t m t s

tm t m t s

ds sm

m ds

x t x v t v

v t v e e

x t x e e sv


Brownian motion and Langevin equation of motion – continued Note that since (t) is a stochastic variable, so is v(t) and x(t)

2

1 2 1 1 2

1

1

21 2

2

( / ) ( / )( )0

0

( / ) ( / )( ) ( / )( )21 2 0 2 2 12

0 0

( / )( / )20

22

0

1

( )

( ) ( )

1 (

( ) x

)

(s )

2 2

tm t m t s

t tm t t m t s m t s

m t tm t t

v t v e e

v

ds sm

g ds ds st v t v e e e

v e e

mx t

m

g gm m

22 ( / ) ( / )02 22

1 1m t m tg mgm

v e t e


Brownian motion and Langevin equation of motion – continued Note that since (t) is a stochastic variable, so is v(t) and x(t)

1 21 2 ( / )( / )21 2 0

2 22 2 ( / ) ( / )0 02 2

20

( ) ( )

( ) x 1 1

Under conditions of thermal equilibrium, argue that

2

2 2m t tm t t

m t m t

T

v t v t v e e

m g mx t v e t e

kT

g gm m

m

m

v

g

1 2

2 20 0

( / )1 2

2 ( / )0

2

( ) ( )

2

0

( ) x

22

1

T T

m t t

T

m t

T

g g mm

k

v v kT

v t v t e

kT m

T

t

m

x t e


Brownian motion and Langevin equation of motion – continued It is interesting to take the Fourier transform of the correlation function

1 2( / )1 2

( / )1

22

, 1

( ) ( )

( ) ( )

= 2 //

v

m t t

T

miv

i

T

v t v t e

S d v t v t

kTm

kTe d eem

kT mm m

1 1 2

, 1

2

1

Spectral properties of random force:( )

( ) (

0

( ) (

) )

) 2

(

i

T

t

t t g

d t

t t

S e kt g T


Example of Brownian systemConsider a particle of mass m attached to a harmonic spring with spring constant m0

2 constrained to move in one dimension:

20

2 2 20 0 0

( ) 1 ( )( ) ( ) ( )

Assume that the particle is initially in thermal equilibrium

with and T T

dv t dx tv t t v tdt m m dt

kTv x kTm

x

2 20

2( )0 0

00

( ) cosh( ) sinh( )

1( ) ( ) sinh( ) ' ( ) ( '

Solution of equation:

Defining

)

tt t t t

m

C t t t

v t v C t t dt t e C t txe em


Example of Brownian system -- continued

20

( ) 1 ( )( ) ( ) ( ) dv t dx tv t t v tdt m m dt

x

2( )0 0

00

1( ) ( ) sinh( ) ' ( ) ( ')t

t t t tv t v C t txe e dt t e C t tm

2 1

2 1

0 0 2 1

22 1 0 2

4 20 0

1

1

2

0 assuming

( ) ( ) ( ) ( )

sinh( ) s

Computing thermally averaged correlation function, noting

inh

that

(

T

t t

TT

t tT

x t t

v t v t v C t C te

xt

v

e

1

1 2( ) ( )12 2

0

2 )

' ( ') ( ')t

t t t t

t

g dt e e C t t C t tm



2 1

2 1

1

1 2

2 1

22 1 0 2

4 20 0

12

( ) ( )12

0

1

2

2

For:

( ) ( ) ( ) ( )

sinh( ) sinh( )

' ( ') ( ')

As

t t

TT

t tT

tt t t t

t t

v t v t v C t C t

t t

g dt e e C t t C t t

e

x

m

e

11| |

suming that 4 ; after some algebra:

( ) ( ) = cosh | | sinh | |T

g kTkTv t v t em

2 20 Notation:

m



| |11( ) ( ) = cosh | | sinh | |

T

kTv t v t em

2 20 Notation:

m

1

| |

, 1

Spectral function:

( ) ( )

cos h | | sinh | |

v vi

T

i

e

kT e em

S d v t v t

d



1

| |

, 1

Spectral function:

( ) ( )

cos h | | sinh | |

v vi

T

i

e

kT e em

S d v t v t

d

,v vS

0 0

0 0


Probability analysis of Brownian motion Fokker-Planck equationRather than analyzing single Brownian particles, it is convenient

to analyze their probability density( , , , ) : probability of finding the Brownian particle at time with

positionx v t t

between and velocity between and random force Averaging over random forces:

( , , ) ( , , , )

x x dxv v dv

P x v t x v t

2

2 2

1 ( )2

( ) Force on particle due to

Fokker-Planck equat

potential

i

(

o

)

n:

P P g Pv v F x Pt x v m m m v

VF x V xx

1 1 22

g comes from random force:( ) ( ) ( )t tgt t


Probability analysis of Brownian motion Fokker-Plank equationJustification of Fokker-Plank equation

x

vS

( , , , )x v t

( , , , ) ( , , , )( , , , )

( , , , ) ( , ,

Continuity co

, ) ( , , ,

n i

)

d tion

x x v t v x v tdxdv x v t dxdv

t x v

x v t x x v t v x v tt x v



(Langevin equation in presence of friction ( ) and potential force ( )) ( )

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

:

F x xdv t dx tv t F x t v tdt m m m dt

V

( , , , ) ( , , , ) ( , , , )

becomes:

1 1( ) ( ) ( )

1 1

Continuity condit

= (

ion

) ( )

x v t v x v t v x v tt x v

vv t F x t

t x v m m m

v vF x t

x m v m v m v



0 1

1

0

1 1 = ( ) ( )

( ) ( )1where ( )

1 ( )

v vF x t

t x m v m v m vL t L t

L v v F xx m m v m v

L tm v

0

0 01

0

( ) ( )

( )

De

( )( ) ( )

Formal solution for (

fine:

( )

) : ( ) exp ' ( ') (0)

L t

L t L t

t

t e t

V t e L t et tt

t t dt t

t

V

V



0

0

0

00

( ) ( )

( ) exp ' ( ') (0)

Evaluation of equation using identity:

( ) ' ( '

!

( 1)!

) (0)

L t

t

x

nt

n

n

n

n

t e t

t dt V t

e

t dt

xn

Vn

t

0 0

2

00

1 11( ) ( ) ( )

Averaging ove

Recall:

r random force:

1( ) ' ( ') (0)

2 !

L t L t

nt

n

V t e L t e L tm v

t dt V tn



00 0

0 0

' '2

0 0 0 0

20

Further analysis:

( ) exp '' ' ( '') ( ') (0)12

12

'' ' ( '') ( ') '

' ' ( t'

) e e e2

''e2

=

t t

t t t tL t tL t L t

tL

t dt dt V t V t

gdt dt V t V t dt dt tm v v

g dtm

0 0

2''

2 et L t

v

0 0

2

2 2

Differential equation for ( )

( )e e (

2

:

)L t L t

t

t g tt m v



0 0

2

2 2

Differential equation for ( )

( )e e (

2

:

)L t L t

t

t g tt m v

0

2

0 2 2

Recall that: ( ) e ( )

( ) ( )( )

:

2

L tt t

t tgL tt m v

Recall: ( , , ) ( , , , )P x v t x v t

2

2 2

Fokker-Planck equation

1 (2

:

)P P g Pv v F x Pt x v m m m v



2

2 2

2

1 ( )2

Define probability c

F

urrent:1ˆ ˆ ( )

2

ˆ ˆ

okker-Planck equati

w r

on:

he e

P P g Pv v F x Pt x v m m m v

g PvP vP F x Pm m m v

Pt x v

J x v

J x v

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