Lecture 19 Boltzmann Statistics.pptx

1
Fisika Termal – PAP3109 Click to edit Master text styles
Jurusan Fisika, Fakultas MIPA – Universitas
Termodinamika dan Fisika Statistik –
FIS62113
Boltzmann Statistics

2
2
Termodinamika dan Fisika Statistik – FIS62113
Boltzmann Statistics
We have followed the following logic:
1. Statistical treatment of isolated systems: multiplicity  entropy  the 2nd Law.
2. Thermodynamic treatment of systems in contact with the heat reservoir  the minimum
free energy principle. However, the link between G and the process of counting of accessible
microstates was not straightforward.
Now we want to learn how to “statistically” treat a system in contact with a heat bath.
The fundamental assumption states that a closed (isolated) system visits every one of its
microstates with equal frequency: all allowed states of the system are equally probable. This
statement applies to the combined system (the system of interest + the reservoir). We wish to
translate this statement into a statement that applies to the system of interest only. Thus, the
question: how often does the system visit each of its microstates being in the thermal
equilibrium with the reservoir? The only information we have about the reservoir is that it is at
the temperature T.
Combined system
U0 = const
Reservoir R
U0 - 
System S

a combined (isolated) system = a heat reservoir and a system in thermal contact

3
3
The Fundamental Assumption for an Isolated System
The ergodic hypothesis: an isolated system in thermal equilibrium,
evolving in time, will pass through all the accessible microstates
states at the same recurrence rate, i.e. all accessible microstates are
equally probable.
The average over long times will equal the average over the ensemble of all equi-energetic
microstates: if we take a snapshot of a system with N microstates, we will find the system in
any of these microstates with the same probability.
Isolated  the energy is conserved. The ensemble of all equi-
energetic states  a mirocanonical ensemble
 1
 2
 i
The probability of a certain macrostate is determined by how many microstates correspond
to this macrostate – the multiplicity of a given macrostate macrostate 
Note that the assumption that a system is isolated is important. If a system is coupled to a
heat reservoir and is able to exchange energy, in order to replace the system’s trajectory by
an ensemble, we must determine the relative occurrence of states with different energies.
Probability of a particular microstate of a microcanonical ensemble
= 1 / (# of all accessible microstates)
Probability of a particular macrostate
= ( of a particular macrostate) / (# of all accessible microstates)

4
4
Systems in Contact with the Reservoir
Reservoir R
U0 - 
System S

We ask the following question: under conditions of equilibrium between the system and
reservoir, what is the probability P(k) of finding the system S in a particular quantum state
k of energy k? We assume weak interaction between R and S so that their energies are
additive. The energy conservation in the isolated system “system+reservoir”:
U0 = UR+ US = const
According to the fundamental assumption of thermodynamics, all the states of the
combined (isolated) system “R+S” are equally probable. By specifying the microstate of the
system k, we have reduced S to 1 and SS to 0. Thus, the probability of occurrence of a
situation where the system is in state k is proportional to the number of states accessible
to the reservoir R . The total multiplicity:
The system – any small macroscopic or microscopic object. If the
interactions between the system and the reservoir are weak, we
can assume that the spectrum of energy levels of a weakly-
interacting system is the same as that for an isolated system.
Example: a two-level system in thermal contact with a heat bath.
         
k
R
k
R
k
R
k
S
k
k U
U
U
U 




 












 0
0
0
0 1
,
R
S
1
2

5
5
Systems in Contact with the Reservoir (cont.)
     
,
1 2
,
0
0 2 0 1
,
1
R
R R
R R
R
R R R
V N V V N
N
S S
T
S U U
U U T
S
U
U
 
 
   
  

 


 



    
  
 
    
       
0
,
0
0 U
U
U
U
S
U
S
U
S
N
V
R
R
R
R 










SR
UR
U
0
-

1
U
0
-

2
S(U0- 2)
S(U0- 1)
S(U0)
U
0
     
N
V
R
R
i
R
i
R U
dU
dS
U
S
U
S
,
0
0
0 







 

Let’s now use the fact that S is much smaller than R (US=k << UR).
The ratio of the probability that the system is in quantum state 1 at energy 1 to the
probability that the system is in quantum state 2 at energy 2 is:
 
 
 
 
 
 
 
 
   







 






 











B
R
B
R
R
B
R
B
R
R
R
S
S
k
S
k
U
S
U
S
k
U
S
k
U
S
U
U
P
P
exp
exp
/
exp
/
exp 2
0
1
0
2
0
1
0
2
0
1
0
2
1 







 
R
R
R
R
R dN
PdV
dU
T
dS 



1 (Pr. 6.9 addresses the case when
the 2nd is not negligible)
0
Also, we’ll consider the case of fixed volume and number of
particles (the latter limitation will be removed later, when we’ll
allow the system to exchange particles with the heat bath

6
6
Boltzmann Factor
 
 



















T
k
T
k
P
P
B
B
S
S
2
1
2
1
exp
exp




This result shows that we do not have to know anything about
the reservoir except that it maintains a constant temperature T !
The corresponding probability distribution is known as the
canonical distribution. An ensemble of identical systems all of
which are in contact with the same heat reservoir and
distributed over states in accordance with the Boltzmann
distribution is called a canonical ensemble.
 
  






 









 

T
k
k
S
P
P
B
B
R
S
S 2
1
2
1
exp
exp




T is the characteristic of the heat reservoir
The fundamental assumption for an isolated system has been transformed into the
following statement for the system of interest which is in thermal equilibrium with the
thermal reservoir: the system visits each microstate with a frequency proportional to the
Boltzmann factor.
Apparently, this is what the system actually does, but from the macroscopic point of view of
thermodynamics, it appears as if the system is trying to minimize its free energy. Or
conversely, this is what the system has to do in order to minimize its free energy.
exp(- k/kBT) is called the Boltzmann factor
reservoir
T

7
7
One of the most useful equations in TD+SP...
Problem 6.13. At very high temperature (as in the very early universe), the proton and
the neutron can be thought of as two different states of the same particle, called the
“nucleon”. Since the neutron’s mass is higher than the proton’s by m = 2.3·10-30 kg, its
energy is higher by mc2. What was the ratio of the number of protons to the number
of neutrons at T=1011 K?
 
 
  86
.
0
K
10
J/K
10
38
.
1
m/s
10
3
kg
10
3
.
2
exp
exp
exp
exp
11
23
2
8
30
2
2
2























 




















 

T
k
mc
T
k
c
m
T
k
c
m
p
P
n
P
B
B
p
B
n
 
  






 









 

T
k
k
S
P
P
B
B
R 2
1
2
1
exp
exp




Secondly, what matters in determining the ratio of the occupation
numbers is the ratio of the energy difference  to kBT. Suppose
that i = kBT and j = 10kBT . Then (i - j ) / kBT = -9, and
 
  8000
9

 e
P
P
j
i


The lowest energy level 0 available to a system (e.g., a molecule) is
referred to as the “ground state”. If we measure all energies relative
to 0 and n0 is the number of molecules in this state, than the
number molecules with energy  > 0
Firstly, notice that only the energy difference  = i - j
comes into the result so that provided that both
energies are measured from the same origin (reference
energy) it does not matter what that origin is.
 










T
k
n
n
B


exp
0

8
8
More problems
A system of particles are placed in a uniform field at T=280K. The particle concentrations
measured at two points along the field line 3 cm apart differ by a factor of 2. Find the force F
acting on the particles.
Answer: F = 0.9·10-19 N
A mixture of two gases with the molecular masses m1 and m2 (m2 >m1) is placed in a very tall
container. The container is in a uniform gravitational field, the acceleration of free fall, g, is
given. Near the bottom of the container, the concentrations of these two types of molecules
are n1 and n2 respectively (n2 >n1) . Find the height from the bottom where these two
concentrations become equal. Answer:  
 g
m
m
n
n
T
k
h B
1
2
1
2 /
ln


Problem 6.14. Use Boltzmann factors to derive the exponential formula for the density of an
isothermal atmosphere.
The system is a single air molecule, with two states: 1 at the sea level (z = 0), 2 – at a height z.
The energies of these two states differ only by the potential energy mgz (the temperature T
does not vary with z):
 
  

























 





















T
k
mgz
T
k
T
k
mgz
T
k
T
k
P
P
B
B
B
B
B
exp
exp
exp
exp
exp
1
1
1
2
1
2






  









T
k
mgz
z
B
exp
0


At home:

9
9
The Partition Function
The partition function Z is called “function” because it depends on T, the spectrum (thus, V), etc.
 
T
k
T
k
P
B
B












exp
T1
T2

0
(kBT2)-1
(kBT1)-1
The areas under these curves must be
the same (=1). Thus, with increasing T,
1/Z decreases, and Z increases. At T =
0, the system is in its ground state
(=0) with the probability =1.
For the absolute values of probability (rather
than the ratio of probabilities), we need an
explicit formula for the normalizing factor 1/Z:
   
i
B
i
i
Z
T
k
Z
P 

 










 exp
1
exp
1
The quantity Z, the partition function, can be found from the normalization condition - the total
probability to find the system in all allowed quantum states is 1:
   

 


i
i
i
i
Z
P 
 exp
1
1 or
   
 

i
i
N
V
T
Z 
exp
,
, The Zustandsumme
in German
Example: a single particle,
continuous spectrum.
T
kB
1

 we will often use this notation
    T
k
dx
x
T
k
d
T
k
T
Z B
B
B

















0
0
exp
exp 


10
10
Average Values in a Canonical Ensemble
     



i
i
i
i
i P
x
x
x 


We have developed the tools that permit us to calculate the average value of different
physical quantities for a canonical ensemble of identical systems. In general, if the systems in
an ensemble are distributed over their accessible states in accordance with the distribution
P(i), the average value of some quantity x (i) can be found as:
The average values are additive. The average total
energy Utot of N identical systems is:
U
N
Utot 
 
exp exp
1
i
i i
i i
B B
k
k
Z
T T

  
 

  


     
 
 




 

 
 
Another useful representation for the
average energy (Pr. 6.16):
Let’s apply this result to the average (mean) energy of the systems in a canonical ensemble
(the average of the energies of the visited microstates according to the frequency of visits):
 
 
  








  T
k
x
N
V
T
Z
x
B
i
i
i
i
i


 exp
,
,
1
In particular, for a
canonical ensemble:
     


 




i
i
i
i
i
i
i
i
i
Z
Z
P
U 




 exp
1
exp
1
 
Z
T
T
k
Z
Z
Z
U B ln
ln
1 2













thus, if we know Z=Z(T,V,N),
we know the average energy!

11
11
Example: energy and heat capacity of a two-level system
- lnni
Ei
the slope ~ T
1= 0
2= 































T
k
T
k
T
k
Z
B
B
B


exp
1
exp
0
exp
The average energy:
 






































T
k
T
k
T
k
T
Z
U
B
B
B 



exp
1
exp
0
exp
0
1
The partition
function:
T
 /2
0 The heat capacity at constant volume:
 
   
 
 2
2
/
exp
1
/
exp
/
/
exp
1 T
k
T
k
T
k
k
T
k
T
T
U
C
B
B
B
B
B
V
V


























CV
(check that the same result follows from )





Z
Z
U
1
U

12
12
 
exp
1 0
0 exp exp
1 exp
exp
exp
1 1
1 1
exp 1 exp
1 exp exp
exp exp
B
B B
B
B
B
B B
B B
B B
k T
U
Z T k T k T
k T
k T
k T
k T k T
k T k T
k T k T








  
 
 
 
 

 
 
     
      
 
   
 
   
   
 
 
 

 
 
 

 
 
   
  
 
 
     
   
    
       
 
   
   

 
 
 
 
 
 
 
 
 
 
 
 
 
 
2 2
2
2
2 2
2 2
1 exp / /
0 exp /
1 exp / 1 exp / 1 exp /
exp / exp /
exp /
1 exp / 1 exp / 1 exp /
B B
B
V
V B B B
B B B
B
B B
B
B B
k T k T
k T
U T T
C
T T k T k T k T
k T k k T
k T
k T k T
k T
k T k T
 
  

  
 

  

  
  
 
 
 
   
   
 
 
  
       
 
     
   
 
   
   
  
   
  
     
2
B
k T
 
 

13
13
Partition Function and Helmholtz Free Energy
Now we can relate F to Z: 






 



















T
k
F
T
T
k
T
F
T
T
F
T
T
F
T
F
TS
F
U
B
B
2
2
2 1
Comparing this with the expression for
the average energy:
 
F
T
k
F
Z
B












 exp
exp
This equation provides the connection between the microscopic world which we
specified with microstates and the macroscopic world which we described with F.
If we know Z=Z(T,V,N), we know everything we want to know about the thermal
behavior of a system. We can compute all the thermodynamic properties:
N
V
B
B
N
V
T
Z
T
k
Z
k
T
F
S
,
,
ln
ln 























N
T
B
N
T
V
Z
T
k
V
F
P
,
,
ln























V
T
B
V
T
N
Z
T
k
N
F
,
,
ln

























14
14
 
 
   
exp exp
ln ln exp
1
ln
ln
ln ln ln
ln
ln
B
B B B
B
F
Z F
k T
Z F F
F Z
F Z
k T Z k T Z k Z T
T T T T
F Z
S k Z T
T T

 

 
   
 
 
   
 
   
 
      
 
   
 
 
 
   
 
 
 
,
( ln ) (ln )
B
B
T N
k T Z
F Z
P k T
V V V
   
 
    
 
  
 

15
15
Microcanonical  Canonical
Microcanonical ensemble Canonical ensemble
  
 ln
,
, B
k
N
V
U
S   Z
T
k
N
V
T
F B ln
,
, 



1
n
P
T
k
E
n
B
n
e
Z
P


1
 the probability of finding a
system in one of the accessible
states
 the probability of finding a
system in one of these states
For the canonical ensemble, the role of Z is similar to that of the multiplicity  for the
microcanonical ensemble. This equation gives the fundamental relation between statistical
mechanics and thermodynamics for given values of T, V, and N, just as S = kB ln gives the
fundamental relation between statistical mechanics and thermodynamics for given values of
U, V, and N.
Our description of the microcanonical and canonical ensembles was based on counting the
number of accessible microstates. Let’s compare these two cases:
 For an isolated system
 the multiplicity  provides the number of
accessible microstates. The constraint in
calculating the states: U, V, N – const
 For a fixed U, the mean temperature T is
specified, but T can fluctuate.
 For a system in thermal contact with reservoir,
 the partition function Z provides the # of
accessible microstates. The constraint: T, V, N
– const
 For a fixed T, the mean energy U is specified,
but U can fluctuate.
 in equilibrium, S reaches a maximum  in equilibrium, F reaches a minimum
V
U
N
U
N
V
N
S
T
V
S
T
P
U
S
T ,
,
,
1



































V
T
N
T
N
V
N
F
V
F
P
T
F
S
,
,
,



































 

16
16
Boltzmann Statistics: classical (low-density) limit
Violations of the Boltzmann statistics are observed if the the density of particles is very large (neutron
stars), or particles are very light (electrons in metals, photons), or they are at very low temperatures
(liquid helium),
However, in the limit of small density of particles, the distinctions between Boltzmann, Fermi, and Bose-
Einstein statistics vanish.
strong exchange interaction weak exchange interaction
  3
/
1
n


the de Broglie wavelength the average distance btw particles
m
T
k
h
p
h
B
~

 (for N2 molecule,  ~ 10-11 m at RT)
  1
3
/
1

m
T
k
n
h
B
Boltzmann
statistics
applies
We have developed the formalism for calculating the thermodynamic properties of the
systems whose particles can occupy particular quantum states regardless of the other particles
(classical systems). In other words, we ignored all kind of interactions between the particles.
However, the occupation numbers are not free from the rule of quantum mechanics. In
quantum mechanics, even if the particles do not interact through forces, they still might
participate in the so-called exchange interaction which is dependent on the spin of interacting
particles (this is related to the principle of indistinguishability of elementary particles, we’ll
consider bosons and fermions in Lecture 23). This type of interactions becomes important if
the particles are in the same quantum state (the same set of quantum numbers), and their
wave functions overlap in space:

17
17
Degenerate Energy Levels
If several quantum states of the system (different sets of quantum numbers) correspond to the
same energy level, this level is called degenerate. The probability to find the system in one of
these degenerate states is the same for all the degenerate states. Thus, the total probability to
find the system in a state with energy i is
Example: The energy levels of an electron in the hydrogen atom:
  









T
k
d
P
B
i
i
i

 exp
  2
6
.
13
i
i
n
n
eV



where ni = 1,2,... is the principle quantum number (these levels are
obtained by solving the Schrödinger equation for the Coulomb
potential). In addition to ni, the states of the electron in the H atom are
characterized with three other quantum numbers: the orbital quantum
number li max = 0,1, ..., ni – 1, the projection of the orbital momentum mli
= - li, - li+1,...0, li-1,li, and the projection of spin si = ±1/2. In the absence
of the external magnetic field, all electron states with the same ni are
degenerate (the property of Coulomb potential). The degree of
degeneracy in this case:
    2
1
0
2
2
1
1
2
1
1
2
2
max
min
i
i
i
n
l
l
i
i n
n
n
l
d
i






 




1
2
i
r
d1 =2
d2 =8
(for a continuous
spectrum, we need
another approach)
di =2ni
2
where di is the degree of degeneracy.
 









i B
i
i
T
k
d
Z

exp
Taking the degeneracy of energy levels into account, the
partition function should be modified:

18
18
Problem (final 2005, partition function)
Consider a system of distinguishable particles with five microstates with energies 0, , , , and
2 (  = 1 eV ) in equilibrium with a reservoir at temperature T =0.5 eV.
1. Find the partition function of the system.
2. Find the average energy of a particle.
3. What is the average energy of 10 such particles?
424
.
1
018
.
0
406
.
0
1
2
exp
exp
3
1
1 
























T
k
T
k
Z
B
B


the average energy of a single particle:
 
2
3exp 2 exp
0.406 0.036
1 eV 0.310 eV
1.424
2
1 3exp exp
B B
i i
i
B B
k T k T
P
k T k T
 
 
  
 
   
    
   

   
    
   
   
   
   

the average energy of N = 10 such particles: 10 10 0.310 eV 3.1 eV
U N U
   
the same result
you’d get from this:
       
   








2
exp
exp
3
1
2
2
exp
exp
3
1


















Z
Z

19
19
Problem 1 (partition function, average energy)
Consider a system of N particles with only 3 possible energy levels separated by  (let the ground state
energy be 0). The system occupies a fixed volume V and is in thermal equilibrium with a reservoir at
temperature T. Ignore interactions between particles and assume that Boltzmann statistics applies.
a. (2) What is the partition function for a single particle in the system?
b. (5) What is the average energy per particle?
c. (5) What is probability that the 2 level is occupied in the high temperature limit, kBT >> ? Explain
your answer on physical grounds.
d. (5) What is the average energy per particle in the high temperature limit, kBT >> ?
e. (3) At what temperature is the ground state 1.1 times as likely to be occupied as the 2 level?
f. (25) Find the heat capacity of the system, CV, analyze the low-T (kBT<<) and high-T (kBT >> ) limits,
and sketch CV as a function of T. Explain your answer on physical grounds.
(a)   

 2
1
exp 





  e
e
d
Z
i
i
i
(b)    












 2
2
2
2
1
2
1
2
1























e
e
e
e
e
e
e
e
Z
Z
(c)
3
1
2
1
1
1
2
1
1 2
2









 








e
e
e
P all 3 levels are populated with
the same probability
(d) 


 











 



1
1
1
2
1
1
2
2
2
e
e
e
e

20
20
Problem 1 (partition function, average energy)
(e)  
1
.
1
ln
2
1
.
1
ln
2
1
.
1
1
2
exp
B
k
T


 



(f)
        
 
 
     
 
 
 2
2
3
2
2
2
2
2
4
3
2
4
3
3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
4
1
4
4
4
4
4
1
2
2
1
4
1
2
2
1
2
2
1












































































































































































e
e
e
e
e
T
k
N
e
e
e
e
e
e
e
e
e
e
e
T
k
N
e
e
e
e
e
e
e
e
e
e
T
k
N
e
e
e
e
e
e
e
e
e
e
T
k
N
dT
d
d
d
N
dT
d
N
dT
dU
C
B
B
B
B
V
Low T (>>):
 
T
k
B
B
V
B
e
T
k
N
e
e
e
e
e
T
k
N
C







 










 2
2
2
2
3
2
2
2
1
4
  2
2
2
2
3
2
2
2
3
2
1
4
T
k
N
e
e
e
e
e
T
k
N
C
B
B
V


















high T (<<):
T
CV

21
21
Problem (the average values)
A gas is placed in a very tall container at the temperature T. The
container is in a uniform gravitational field, the acceleration of free fall,
g, is given. Find the average potential energy of the molecules.
h
0
H
# of molecules within dh:   dh
area
T
k
mgh
n
dh
h
h
dN
B
)
(
exp
0 











   
 
 
 
0
0 0 0
0
0
0 0
exp ( ) exp exp
exp ( )
exp exp
B B
B B
mgH mgH
k T k T
H
B
B
B
B
mgH mgH
H
B k T k T
B
B
mgh k T
mghn area dh k T y y dy y y dy
k T mg
mgh
U y k T
k T
mgh
n area dh k T
k T y dy y dy
mg
 
  
 
 
 
     
 
   

 
 
 
  

 
       
           
0 0
0
0 0 0
exp exp exp exp
exp exp exp exp exp 1 exp
A A
A A A
A
d
y y dy y y y y y dy
dy
y y y dy y y dy y y dy A A A
 
      
 
 
           
 
 
  
    1
exp
exp
0





 A
dy
y
A
T
k
U B

For a very tall container (mgH/kBT ):

22
22
Partition Function for a Hydrogen Atom (Pr. 6.9)
(a) Estimate the partition function for a hydrogen atom at T =
5800K (= 0.5 eV) by taking into account only three lowest
energy states.

r
1= -13.6 eV
2= -3.4 eV
3= -1.5 eV
Any reference energy can be chosen. Let’s choose  = 0 in the ground state: 1=0,
2=10.2 eV, 3=12.1 eV, etc. The partition function:
1
10
8
.
2
10
5
.
5
1
9
4
exp
9
exp
4
exp 10
9
2
.
24
4
.
20
0
3
2
1










































 



e
e
e
T
k
T
k
T
k
Z
B
B
B



However, if we take into account all discrete levels, the full partition function diverges:
































 




 1
2
1
2
2 6
.
13
exp
1
1
6
.
13
exp
n
B
n B
n
T
k
T
k
n
n
Z
 

 




 












i B
i
i
i B
i
i
T
k
n
eV
n
T
k
d
Z
2
2 /
1
1
6
.
13
exp
2
exp

we can forget about the spin degeneracy – it is the same
for all the levels – the only factor that matters is n2

23
23
Partition Function for a Hydrogen Atom (cont.)
Intuitively, only the lowest levels should matter at  >> kBT . To resolve this paradox,
let’s go back to our assumptions: we neglected the term PdV in
 
R
R
R PdV
dU
T
dS 

1
If we keep this term, then 






 


T
k
PV
E
B
exp
factor
Boltzmann
New
For a H atom in its ground state, V~(0.1 nm)3, and at the atmospheric pressure,
PV~ 10-6 eV (negligible correction). However, this volume increases as n3 (the
Bohr radius grows as n), and for n=100, PV is already ~1 eV. The PV terms cause
the Boltzmann factors to decrease exponentially, and this rehabilitates our
physical intuition: the correct partition function will be dominated by just a few
lowest energy levels.

Lecture 19 Boltzmann Statistics.pptx

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Lecture 19 Boltzmann Statistics.pptx