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Boltzmann Distribution and Helmholtz Free Energy
The Boltzmann factor
- We consider a thermodynamic system (S) in contact
with a Reservoir (R) (Heat Bath), which has
temperature T. S+R is an isolated system with U0
internal energy. We assume S<<R
- central problem: what is the probability that S will be is a given state :s (S is not an isolated
system, and we cannot use the Ps=1/g results)?
- Since S+R is an isolated system, and thus we can
apply our fundamental assumption for the probability of
the states.
- When S is in state s with energy s, R will have an
energy UR=U0- s with multiplicity gR(U0- s ).
- applying our fundamental assumption for the total isolated system and the relation between entropy
and multiplicity, for two states (1 and 2) with energies 1 and 2 we get:
P(1 ) g R (U 0  1 ) exp[ R (U 0  1 )]


P( 2 ) g R (U 0   2 ) exp[ R (U 0   2 )]
We assume now i<<U0, and write
 
 (U   )   (U )   

 ....   R (U 0 )   i
R
R
0
i
R
0
i
 U U

 ...
0
P( ) exp( /  )

P( ) exp( /  )
and we get the important relation:
1
1
1
2
2
e



Boltzmann
factor
Z ( )   exp( s /  )
partition function:
{ s}
- The partition function is the proportionality factor between the probability P(s) and
 P( s )  1
Boltzmann factor!
exp (  ε s /τ )
P(ε s ) 
Z
U      s P( s ) 
{ s}
  s exp( s /  )
{ s}
Z
Normalization satisfied!
{ s}
One of the MOST USEFUL and
IMPORTANT results of statistical physics
2
 ln(Z )

<…> --> ensemble average (sometimes called thermal average)
Average energy of the
S thermodynamic
system
Energy and heat capacity of a two state system
We consider a thermodynamic system with two possible states (energies 0 and ) in contact with a
heath-bath at temperature T. We calculate the internal energy and heat capacity of the system as a
function of the system temperature.
Z  exp(0 /  )  exp( /  )  1  exp( /  )
U   
 exp( /  )
Z

exp( /  )
1  exp( /  )
exp( /  )
 U 
 
CV  
  
2
  V    [exp( /  )  1]
2
2
  
exp( / k BT )
 U 


CV  

k

B
2

 T V
 k BT  [exp( / k BT )  1]
for >>
for <<

CV  ( / 2 ) 2
CV   /   exp( /  )
2
0
Pressure
We consider a system in a quantum state s of energy s.
We assume that s depends on volume V (Example
quantum particle closed in a 3d box)
We consider a quasi-static
volume change: V--> V-dV
 s (V  dV )   s (V )  (d s / dV )dV  ...
dU  U (V  dV )  U (V )  (d s / dV )dV  ps dV
 U 
p  

 V  σ
calculation of pressure (during the quasi-static and reversible process
the total number of micro-states were kept constant--> entropy was kept
constant)
  
  
  
  
d (U ,V )  
 (U )  
 (V )
 dU  
 dV for “isentropic” process: 0  
 U V
 V U
 U V
 V U
 σ 
p  τ

 V U
    U 
  

 
  


U

V

V

V 


U
   (U )   
0



 U V (V )  V U
Thermodynamic identity
The combined mathematical form of the first and second law of thermodynamics (when the
particle number is fixed)


1
p




d (U ,V )  
 dU  
 dV  dU  dV

U

V



V

U
dU  τdσ  pdV
dU  TdS  pdV
Problems:
For a binary model system formed by 4 spins with magnetic moments +/- m in magnetic field B
and in contact with a reservoir at temperature T find:
1. The probability that the total magnetization is 4m.
2. The probability that the total magnetization is 0. At which temperature will this probability
become 1?
3. The probability that one randomly selected spins is pointing in the + direction.
4. The average energy of the system.
Extra problem:
Consider a binary model system formed by 4 spins with magnetic moments +/- m in magnetic
field B and in contact with a reservoir at temperature T. By using the Renyi entropy formula
(see the extra problem from the end of lecture-note 3) , calculate the entropy at an arbitrary
temperature T. Generalize the result for N spins.