Thermo & Stat Mech - Spring 2006 Class 27 1 Thermodynamics and Statistical Mechanics Random Walk

Thermo & Stat Mech - Spring 2006 Class 27

1

Thermodynamics and Statistical Mechanics

Random Walk


2

Random Walk

Often called “drunkard’s walk”. Steps in random directions, but on average, how far does he move, and what is the standard deviation? Do for one dimension. Consider each step is of length s0, but it can be either forward or backward.

Probability of going forward is p.


3

Random Walk

After N steps, if n are forward, the distance traveled is,

S = [n – (N – n)]s0 = (2n – N) s0

The probability of this occurring is,

nNnqpnNn

NnP

)!(!!

)(


4

Random Walk

The average distance covered after N steps is,

0 and ,0 ,For

stepper distance average The )12(

where

)12(

)2()2(

21

0

0

00

Ssp

sps

s NspNS

sNpNsNnS


5

Random Walk

Standard deviation.

step)per (s.d. 4 where

2 4

44

22

220

22

020

2

20

220

2

2

00

22

spqs

sN

NpqspqsN

Npqsnns

sNnsNnSS


6

Random Walk

As before,

NS

Nppq

spNNpqs

S

1

1)12(

2)12(

2

0

0


7

3D Random Walk

Assume the direction of each step is random.

0sin)cos(4

1cos 00

ddsssz

Average distance moved per step is zero.


8

Standard Deviation

220

20

222 cos)cos()()( ssssss zzzz

since sz 0.

0

220

2

00

2202

sincos2

sincos4

)(

ds

dds

sz


9

Standard Deviation

33

2

23

1

3

1

2

3222)(

20

20

20

1

1

320

1

1

220

1

1

2202

sss

xsdxx

sdxx

ssz

Let cos = x, and sin d = – dx. Then,


10

Gaussian Distribution

When dealing with very large numbers of particles, it is often convenient to deal with a continuous function to describe the probability distribution, rather than the binomial distribution. The Gaussian distribution is the function that approximates the binomial distribution for very large numbers.


11

Binomial Distribution

Let us develop a differential equation for P in terms of n, and treat n as continuous. Then we can solve the equation for P.

nNnqpnNn

NnP

)!(!!

)(


12


If n increased by one, then the change in P is

)!()!(!

))(1()!()!(!

)!()!(!

)!1()!1(!

)()1(

1

1

11

nNnqpN

nNnpq

nNnqpN

P

nNnqpN

nNnqpN

P

nPnPP

nNnnNn

nNnnNn


13


1)1()(

)(1))(1(

)(

1))(1()!()!(

!

)!()!(!

))(1()!()!(!

1

1

1

1

1

1

qnpnN

nPnNn

pqnPP

nNnpq

nNnqpN

P

nNnqpN

nNnpq

nNnqpN

P

nNn

nNnnNn


14


NpqNpn

nPnq

nNpnP

qnqnNp

nPP

qnqqpnNp

nPqn

qnqnpNpnPP

qnqnpnN

nPqnpnN

nPP

)()()1(

)(

)1()(

)()1(

)(

)1()1()(

)(1)1()(

)(


15


dnnn

PdP

nnnP

dndP

nnnP

NpqNpn

nPP

2

2

2

)(

)()(


16

Binomial to Gaussian Distribution

2

2

2

)(

2

2

21

2

const.)(

ln

so ,

nn

CeP

nnP

dnnn

PdP


17


What is C?

dxeCdxxP

CexP

dxdnxnn

x

x

2

2

2

2

2

2

1)(

Then,

)(

so , and ,Let


18


02/

022

0 022

02

022

02

022

2

2

2

22

2

2

2

2

2

2

2

2

2

2

41

441

21

21

drdreC

dxdyeCdyedxeC

dyeC

dxeCdxeC

r

yxyx

y

xx


19


21

or ,2

1

221

2441

22

0

2220 2

222

022

02/

022

2

2

2

2

2

2

2

2

CC

eCrdr

eC

rdreCdrdreC

rr

rr


20


2

2

2

2

2

)(

2

21

)(

21

)(

nn

x

enP

exP


21

Properties of Gaussian Distribution

997.0)(

954.0)(

683.0)(

0)(

33

22

dxxP

dxxP

dxxP

nndxxxPx


22

Properties of Gaussian Distribution

22

2

2

)(

at 0)(

)(or 0at 0)(

dxxPx

xx

xP

nnxxxP


23

Problem

A bottle of ammonia is opened briefly. The molecules move s0 = 10-5 m in any direction before a collision. There are 107 collisions per second. How long until 32% of the molecules are 6 m or more from the bottle?


24

Solution

= 6 m

2 2N sz( ) , where N = (107s-1)t

2-172 )()s 10( zst


25

Solution

201-1-7

2

21-7

2

2-172

m 1031

)s 10(

m) 0.6(

)()s 10(

)()s 10(

z

z

st

st

t = 1.08 ×105 s = 30 hr = 1.25 days

Documents

Thermo & Stat Mech - Spring 2006 Class 27 1 Thermodynamics and Statistical Mechanics Random Walk