Thermodynamics and Statistical
Mechanics
Random Walk
Thermo & Stat Mech Spring 2006 Class 27
1
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.
Thermo & Stat Mech - Spring 2006
Class 27
2
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,
N!
n N n
P ( n) 
p q
n!( N  n)!
Thermo & Stat Mech - Spring 2006
Class 27
3
Random Walk
The average distance covered after N steps is,
S  (2n  N ) s0  (2 pN  N ) s0
S  N (2 p  1) s0  Ns
where
s  (2 p  1) s0 The average distance per step
For p  12 , s  0, and S  0
Thermo & Stat Mech - Spring 2006
Class 27
4
Random Walk
Standard deviation.
 

 4 s n  n   4 s Npq

 
  S  S  2n  N s0  2n  N s0
2

2
2
2
2
0
 2  N 4 pqs02
2
2
0
  2 s0 Npq
 2  N s 2
where
4 pqs02  s 2
(s.d. per step)
Thermo & Stat Mech - Spring 2006
Class 27
5
Random Walk
As before,
 2s0 Npq
2 pq 1


S N (2 p  1) s0 (2 p  1) N

1

S
N
Thermo & Stat Mech - Spring 2006
Class 27
6
3D Random Walk
Assume the direction of each step is random.
1
s z  s0 cos  
4
 (s
0
cos  ) sin  d d  0
Average distance moved per step is zero.
Thermo & Stat Mech - Spring 2006
Class 27
7
Standard Deviation
(s z )  ( s z  s z )  s  ( s0 cos  )  s cos 
2
2
since
.
2
z
2
2
0
2
sz  0
2 
0
2
s
2
(s z ) 
cos
 sin  d  d

4 0
0
2
2 
0
s
2
  cos  sin  d
2 0
Thermo & Stat Mech - Spring 2006
Class 27
8
Standard Deviation
Let cos  = x, and sin  d = – dx. Then,
2 1
2
0
s
(s z )  
2
2
s02

2
2 1
0
s
1 x dx  2
2
0
s
1 x dx  2
2
3 1
x 
 
 3  1
2
2
s
s
1

1
2


0
0



 3 3  2 3 3
Thermo & Stat Mech - Spring 2006
Class 27
9
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.
Thermo & Stat Mech - Spring 2006
Class 27
10
Binomial Distribution
N!
n N n
P ( n) 
p q
n!( N  n)!
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.
Thermo & Stat Mech - Spring 2006
Class 27
11
Binomial Distribution
If n increased by one, then the change in P is
P  P (n  1)  P (n)
N ! p n1q N n1
N ! p n q N n
P 

(n  1)!( N  n  1)! (n)!( N  n)!
N ! p n q N n
pq 1
N ! p n q N n
P 

1
(n)!( N  n)! (n  1)( N  n)
(n)!( N  n)!
Thermo & Stat Mech - Spring 2006
Class 27
12
Binomial Distribution
n
N n
1
n
N n
N! p q
pq
N! p q
P 

1
(n)!( N  n)! (n  1)( N  n)
(n)!( N  n)!
N n
1


N! p q
pq
P 
 1
1

(n)!( N  n)!  (n  1)( N  n)

n
1


pq
 ( N  n) p 
P  P (n) 
 1  P (n) 
 1
1
 (n  1)q

 (n  1)( N  n)

Thermo & Stat Mech - Spring 2006
Class 27
13
Binomial Distribution
 ( N  n) p 
 ( N  n) p  (n  1)q 
P  P (n) 
 1  P (n) 

(
n

1
)
q
(
n

1
)
q




 Np  np  nq  q 
 Np  n( p  q )  q 
P  P (n) 
 P ( n) 


(
n

1
)
q
(
n

1
)
q




 Np  n  q 
 Np  n 
 n  Np 
P  P (n) 
 P (n) 
  P (n) 



(
n

1
)
q
nq
Npq






Thermo & Stat Mech - Spring 2006
Class 27
14
Binomial Distribution
 n  Np 
n  n
P   P (n) 
  P ( n)  2 

 
 Npq 
dP
n  n
  P (n)  2 
dn
 
dP
n  n
   2  dn
P
 
Thermo & Stat Mech - Spring 2006
Class 27
15
Binomial to Gaussian
Distribution
dP
n  n
   2  dn, so
P
 
2
(
n

n
)
1
ln P   2
P  Ce

2
 const.
( nn )2

2 2
Thermo & Stat Mech - Spring 2006
Class 27
16
Gaussian Distribution
What is C?
Let n  n  x, and dn  dx, so
P ( x)  Ce
Then,


x2
 2
2


 P ( x)dx  1  C  e
x2
 2
2
Thermo & Stat Mech - Spring 2006
Class 27
dx
17
Gaussian Distribution


1  C e
x2
 2
2

0
1  2C  e

0
dx  2C  e
y2
 2
2
2 
0
1  4C  e
dx
dy
x2
 2
2

0
dx  e
r2
 2
2   /2
2
0 0
1  4C  
x2
 2
2
e
y2
 2
2
2  
0 0
dy  4C   e
x2  y 2

2 2
dxdy
rddr
Thermo & Stat Mech - Spring 2006
Class 27
18
Gaussian Distribution
r2
 2
2   /2
2
0 0
1  4C  
e
2 
0
1  2C   e
2
C 
2
1
2
2
rddr 4C
r2
 2
2
, or
2

2

0
 e
r2
 2
2

 2C   e
2


1
C
2 
rdr
2
Thermo & Stat Mech - Spring 2006
Class 27
2
rdr
r2
 2
2



 0
19
Gaussian Distribution
1
P( x) 
e
2 
1
P ( n) 
e
2 
x2
 2
2
( nn )2

2 2
Thermo & Stat Mech - Spring 2006
Class 27
20
Properties of Gaussian
Distribution


x   xP( x)dx  0


P ( x)dx  0.683
2
2
P ( x)dx  0.954
3
3
P ( x)dx  0.997


nn
Thermo & Stat Mech - Spring 2006
Class 27
21
Properties of Gaussian
Distribution
P ( x)
 0 at x  0 (or n  n)
x
2
 P( x)
 0 at x  
2
x


2
2
x
P
(
x
)
dx



Thermo & Stat Mech - Spring 2006
Class 27
22
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?
Thermo & Stat Mech - Spring 2006
Class 27
23
Solution
 =6m
  N ( sz )
2
2
, where N = (107s-1)t
  (10 s )t(sz )
2
7
-1
2
Thermo & Stat Mech - Spring 2006
Class 27
24
Solution
 2  (107 s -1 )t(s z ) 2

2
2
(6.0 m)
t

(107 s -1 )(s z ) 2 (107 s -1 ) 1 10-10 m 2 
3

t = 1.08 ×105 s = 30 hr = 1.25 days
Thermo & Stat Mech - Spring 2006
Class 27
25