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Thermodynamic Potentials
We have been using the total internal energy U and, sometimes, the enthalpy H to
characterize various macroscopic systems. These functions are called the
thermodynamic potentials: all the thermodynamic properties of the system can be
found by taking partial derivatives of the TP.
For each TP, a set of so-called “natural variables” exists:
d U  T d S  P dV   d N
d H  T d S  V dP   d N
The other two thermodynamic potentials: the Helmhotz free energy F and Gibbs
free energy G. Depending on the type of a process, one of these four
thermodynamic potentials provides the most convenient description (and is
tabulated). All four functions have units of energy.
Potential
Variables
U (S,V,N)
S, V, N
H (S,P,N)
S, P, N
F (T,V,N)
V, T, N
G (T,P,N)
P, T, N
When considering different types of
processes, we will be interested in two main
issues:
what determines the stability of a system
and how the system evolves towards an
equilibrium;
how much work can be extracted from a
system.
Diffusive Equilibrium and Chemical Potential
For completeness, let’s recall what we’ve learned about the chemical potential.
d U  T d S  P dV   d N
UA, VA, SA
UB, VB, SB
For sub-systems in
diffusive equilibrium:
dS
1
P

d U  dV  d N
T
T
T
The meaning of the partial derivative (S/N)U,V : let’s fix
VA and VB (the membrane’s position is fixed), but
assume that the membrane becomes permeable for gas
molecules (exchange of both U and N between the subsystems, the molecules in A and B are the same ).
 S AB 


0
 U A VA , N A
 S AB 


0

N
A U A , V A

 S A  SB

 N A  NB
 S 
 U 
 

 N U ,V  N  S ,V
  T 
S
In equilibrium,
NA
TA  TB

 S 

 
T
 N U ,V
- the chemical
potential
 A  B
PA  PB
Sign “-”: out of equilibrium, the system with the larger
UA
S/N will get more particles. In other words, particles
will flow from from a high /T to a low /T.
Chemical Potential of an Ideal gas
 S 
 U 





N

N

U ,V 
 S ,V
  T 
 has units of energy: it’s an amount of energy we need to (usually) remove
from the system after adding one particle in order to keep its total energy fixed.
Monatomic ideal gas:
3/ 2

5
   4 m  

5/ 2
S ( N ,V ,U )  N k B ln V 
U

ln
N




2
3
h
2
 


  

3/ 2

 h3

S
V
2

m
P 
 

 

  T     k BT ln   2 k BT    k BT ln 

3/ 2
5/ 2 
 N U ,V
 
 N  h
 2m k BT  
At normal T and P,  for an ideal gas is negative (e.g., for He,  ~ - 5·10-20 J ~ - 0.3 eV).
Sign “-”: by adding particles to this system, we increase its entropy. To keep
dS = 0, we need to subtract some energy, thus U is negative.

The chemical potential increases with with its pressure. Thus,
0
when P
the molecules will flow from regions of high density to regions of
increases
lower density or from regions of high pressure to those of low

pressure .
Note that  in this case is negative because S increases with n. This is not always the
case. For example, for a system of fermions at T0, the entropy is zero (all the lowest
states are occupied), but adding one fermion to the system costs some energy (the Fermi
energy). Thus,
 T  0  EF  0
The Quantum Concentration
3/ 2
3/ 2
2


 V  2 m

 n


h



  k BT ln  
   k BT ln   2 k BT    k BT ln n 

n 
 
  2  mkBT  
 N  h
 Q
where n=N/V is the concentration of particles
When n << nQ (In the limit of low densities), the gas is in the classical regime, and <0.
When n  nQ,   0
 2 m

nQ   2 k BT 
 h

dB
3/ 2
- the so-called quantum concentration (one particle per
cube of side equal to the thermal de Broglie wavelength).
h
h
 
p
mkBT
nQ 
1
dB 3



mkBT 

h2 
At T=300K, P=105 Pa , n << nQ.
When n  nQ, the quantum statistics comes into play.
3/ 2
Isolated Systems, independent variables S and V
Advantages of U : it is conserved for an isolated system (it also has a simple
physical meaning – the sum of all the kin. and pot. energies of all the particles).
In particular, for an isolated system Q=0, and dU = W.
Earlier, by considering the total differential of S as a function of variables U, V, and
N, we arrived at the thermodynamic identity for quasistatic processes :
dU S ,V , N   T dS  PdV  dN
The combination of parameters on the right side is equal to the exact differential of U
. This implies that the natural variables of U are S, V, N,
Considering S, V, and N
as independent variables:
 U 
 U 
 U 
 dS  
 dV  
 dN
dU ( S ,V , N )  

S

V

N

V , N

S ,N

 S ,V
Since these two equations for dU must yield
the same result for any dS and dV, the
corresponding coefficients must be the same:
 U 

  T

S

V , N
 U 

   P

V

S ,N
 U 

  

N

 S ,V
Again, this shows that among several macroscopic variables that characterize the
system (P, V, T, , N, etc.), only three are independent, the other variables can be
found by taking partial derivatives of the TP with respect to its natural variables.
Isolated Systems, independent variables S and V (cont.)
Work is the transfer of energy to a system by a change in the external parameters
such as volume, magnetic and electric fields, gravitational potential, etc. We can
represent W as the sum of two terms, a mechanical work on changing the volume of
a system (an “expansion” work) - PdV and all other kinds of work, Wother
(electrical work, work on creating the surface area, etc.):
 W   PdV   Wother
If the system comprises only solids and liquids, we can usually assume dV  0, and
the difference between W and Wother vanishes. For gases, the difference may be
very significant.
initially, the system is not necessarily in equilibrium
The energy balance for
dU  T dS  PdV   Wother  0 (for fixed N)
an isolated system :
 Wother  PdV  TdS
If we consider a quasi-static process (the system evolves from one equilibrium state to
the other), than, since for an isolated system Q=TdS=0,
 Wother  PdV
Equilibrium in Isolated Systems
UA, VA, SA
UB, VB, SB
For a thermally isolated system Q = 0. If the
volume is fixed, then no work gets done (W = 0)
and the internal energy is conserved:
U  const
While this constraint is always in place, the system might be out of equilibrium (e.g.,
we move a piston that separates two sub-systems, see Figure). If the system is
initially out of equilibrium, then some spontaneous processes will drive the
system towards equilibrium. In a state of stable equilibrium no further spontaneous
processes (other than ever-present random fluctuations) can take place. The
equilibrium state corresponds to the maximum multiplicity and maximum entropy. All
microstates in equilibrium are equally accessible (the system is in one of these
microstates with equal probability).
S eq  max
This implies that in any of these spontaneous processes, the entropy tends to
increase, and the change of entropy satisfies the condition
dS  0
Suppose that the system is characterized by a parameter
x which is free to vary (e.g., the system might consist of
ice and water, and x is the relative concentration of ice).
By spontaneous processes, the system will approach the
stable equilibrium (x = xeq) where S attains its absolute
maximum.
S
xeq
x
Systems in Contact with a Thermal Reservoir
When we consider systems in contact with a large thermal reservoir (a “thermal
bath, there are two complications: (a) the energy in the system is no longer fixed (it
may flow between the system and reservoir), and (b) in order to investigate the
stability of an equilibrium, we need to consider the entropy of the combined system
(= the system of interest+the reservoir) – according to the 2nd Law, this total
entropy should be maximized.
What should be the system’s behavior in order to maximize the total entropy?
For the systems in contact with a eat bath, we need to invent a better, more useful
approach. The entropy, along with V and N, determines the system’s energy U =U
(S,V,N). Among the three variable, the entropy is the most difficult to control (the
entropy-meters do not exist!). For an isolated system, we have to work with the
entropy – it cannot be replaced with some other function. And we did not want to do
this so far – after all, our approach to thermodynamics was based on this concept.
However, for systems in thermal contact with a reservoir, we can replace the
entropy with another, more-convenient-to-work-with function. This, of course, does
not mean that we can get rid of entropy. We will be able to work with a different
“energy-like” thermodynamic potential for which entropy is not one of the natural
variables.
Review:
Entropy
Units: J/K
of a system in a given
macrostate (N,U,V...):
S  kB ln  N, U,V ... 
The entropy is a state function, i.e. it depends on the
macrostate alone and not on the path of the system to this
macrostate.
Entropy is just another (more convenient) way of talking about multiplicity.
Convenience:
reduces ridiculously large numbers to manageable numbers
23
Examples: for N~1023,  ~ 1010 , ln  ~1023, being multiplied by kB ~ 10-23, it gives
S ~ 1J/K.
The “inverse” procedure: the entropy of a certain macropartition is 4600kB.
What is the multiplicity of the macropartition?
e
S
kB
 e4600 ~ 102000
if a system contains two or more interacting sub-systems having their own
distinct macrostates, the total entropy of the combined system in a given
macropartition is the sum of the entropies of the subsystems they have in that
macropartition:  AB =  A x  B x  C x....  S AB = S A + S B + S C + ...
Problem:
Imagine that one macropartition of a combined system of two Einstein
solids has an entropy of 1 J/K, while another (where the energy is more
evenly divided) has an entropy of 1.001 J/K. How many times more likely
are you to find the system in the second macropartition compared to the
first?
0.724641023
Prob(mp2)  2 e
e
7.21019

 S1 / k B  0.725361023  e
Prob(mp1) 1 e
e
S2 / kB
Microcanonical  Canonical
Our description of the microcanonical and canonical ensembles was based on counting
the number of accessible microstates. Let’s compare these two cases:
microcanonical ensemble
canonical ensemble
For an isolated system, the multiplicity
 provides the number of accessible
microstates.
The
constraint
in
calculating the states: U, V, N – const
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 U, the mean temperature T
is specified, but T can fluctuate.
For a fixed T, the mean energy U is
specified, but U can fluctuate.
1
Pn 

- the probability of finding
a system in one of the
accessible states
S U ,V , N   kB ln 
- in equilibrium, S reaches a maximum
E
n
1  k B T - the probability of finding
Pn  e
a system in one of these
Z
states
F T ,V , N   kB T ln Z
- in equilibrium, F reaches a minimum
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.
1  S 


T   U V , N
P S 


T   V U , N

 S 

 
T

N

U ,V
F 

S  

T

V , N
F 

P  

V

T , N
F 


N

T ,V
  
Boltzmann Statistics
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
The Fundamental Assumption for an Isolated System
i
Isolated  the energy is conserved
1
2
The ensemble of all equi-energetic states 
a microcanonical ensemble
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.
Probability of a particular microstate of a microcanonical ensemble
= 1 / (# of all accessible microstates)
The probability of a certain macrostate is determined by how many
microstates correspond to this macrostate – the multiplicity of a
given macrostate 
Probability of a particular macrostate
= ( of a particular macrostate) / (# of all accessible microstates)
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.
Systems in Contact
with the Reservoir
R
2
1
S
Reservoir R
U0 - 
System S

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.
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:
 k ,U 0   k   S  k   R U 0   k   1  R U 0   k    R U 0   k 
Systems in Contact with the Reservoir (cont.)
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:
 S U     S R U 0   2 
 S R 
PS 1   R U 0  1  exp S R U 0  1  / k B 




 exp  R 0 1

exp

PS  2   R U 0   2  exp S R U 0   2  / k B 
k
k
B


 B 
SR
Let’s now use the fact that S is much smaller than R (US=k << UR).
S(U0)
S(U0- 1)
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
U0
U0- 1
U0- 2
S(U0- 2)
UR
dS R 
1
dU R  PdVR  dN R 
TR
0
 S

S R U   S R U 0    R U 0  U  U 0 
 U R
V , N
 dS

S R U 0   i   S R U 0    i  R U 0 
 dU R
V , N
 S
 S R

 S R


1
R
U 0   2  
U 0  1  
U 0      1   2
S R  
TR
 U R
 U R
V , N
 U R
V , N
V , N TR 
Boltzmann Factor
 S 
   
PS 1 
 exp  R   exp   1 2 
PS  2 
 kB 
 k BT 
T is the characteristic of the heat reservoir
exp(- k/kBT) is called the Boltzmann factor
  
exp   1 
PS  1 
 k BT 

PS  2 
  
exp   2 
 k BT 
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.
reservoir
T
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.
One of the most useful equations in
Termodynamics + Statistical Physics
 S 
   
P1 
 exp  R   exp   1 2 
P 2 
 kB 
 k BT 
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 it does not matter what that origin
is.
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
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
P i  9
 e  8000
P j 
  
n 

 exp  
n0
k
T
 B 
Helmholtz Free Energy (independ. variables T and V)
Let’s do the trick (Legendre transformation) again, now to exclude S :
U S ,V   U S ,V   T S
F U T S
d U  TS   TdS  PdV  SdT  TdS  SdT  PdV
The natural variables for F are T, V, N:
Comparison yields the relations:
Helmholtz free energy
dF T ,V , N   SdT  PdV  dN
 F 
 F 
 F 
dF T ,V , N   
 dT  
 dV  
 dN

T

V

N

V , N

T , N

T ,V
 F 

  S
 T V , N
 F 

   P can be rewritten as:
 V T , N
 F 

  P
 V T , N
 F 

 
 N T ,V
 F 
 U 
 S 
P  
  
 T

 V T , N
 V T , N
 V T , N
The first term – the “energy” pressure – is dominant in most solids, the second term
– the “entropy” pressure – is dominant in gases. (For an ideal gas, U does not
depend on V, and only the second term survives).
F is the total energy needed to create the system, minus the heat we can get “for
free” from the environment at temperature T. If we annihilate the system, we
can’t recover all its U as work, because we have to dispose its entropy at a nonzero T by dumping some heat into the environment.
The Minimum Free Energy Principle (V,T = const)
The total energy of the combined system (= the system of interest+the reservoir) is
U = UR+Us, this energy is to be shared between the reservoir and the system (we
assume that V and N for all the systems are fixed). Sharing is controlled by the
maximum entropy principle:
S R s U R ,U s   S R U  U s   S s U s   max
system’s parameters only
Since U ~ UR >> Us
F
U

 S 
S R  s U , U s   S R U    R  U s   S s U s   S R U    s  S s   S R U   s
T
 U 
T

SR+s
Fs
reservoir
+system
Us
loss in SR due to
transferring Us to
the system
dS R  s U ,U s   dS s 
gain in Ss due to
transferring Us to
the system
dU s
dF
1
  dU s  TdS s    s
T
T
T
system
Thus, we can enforce the maximum entropy principle by simply
Us minimizing the Helmholtz free energy of the system without
stable
having to know anything about the reservoir except that it
equilibrium
maintains a fixed T! Under these conditions (fixed T, V, and N),
the maximum entropy principle of an isolated system is transformed into a minimum
Helmholtz free energy principle for a system in thermal contact with the thermal bath.
Processes at T = const
In general, if we consider processes with “other” work:
For the processes at T = const
(in thermal equilibrium with a large reservoir):
dF   SdT  PdV   Wother
dF T   PdV   Wother T
The total work performed on a system at T = const in a reversible process is equal
to the change in the Helmholtz free energy of the system. In other words, for the T =
const processes the Helmholtz free energy gives all the reversible work.
Problem: Consider a cylinder separated into two parts by an adiabatic piston.
Compartments a and b each contains one mole of a monatomic ideal gas, and their
initial volumes are Vai=10l and Vbi=1l, respectively. The cylinder, whose walls allow
heat transfer only, is immersed in a large bath at 00C. The piston is now moving
reversibly so that the final volumes are Vaf=6l and Vbi=5l. How much work is delivered
by (or to) the system?
The process is isothermal :
UA, VA, SA
UB, VB, SB
For one mole of
monatomic ideal gas:
The work delivered
by the system:
dF T   PdV T
Va f
Vb f
Va i
Vb i
 W   Wa   Wb   dFa   dFb
3

T
V
  RT ln  RT ln  Tf ( N , m) 
T0
V0
2

Vaf
Vbf
 W  RT ln
 RT ln
 2.6 103 J
Vai
Vbi
3
F  U  TS  RT
2
Gibbs Free Energy (independent variables T and P)
Let’s do the trick of Legendre transformation again, now to exclude both S and V :
U S ,V   U T , P  T S  PV
G  U  T S  PV
- the thermodynamic potential G is called the Gibbs free energy.
Let’s rewrite dU in terms of independent variables T and P :
dU  TdS  PdV  d (TS )  SdT  d PV   VdP
d U  TS  PV   SdT  VdP
dGT , P, N   SdT  VdP  dN
Considering T, P, and N as
independent variables:
 G 
 G 
 G 
dGT , P, N   
 dT  
 dP  
 dN

T

P

N

 P, N

T , N

T , P
 G 
  S
Comparison yields the relations: 

T

 P, N
 G 

 V

P

T , N
 G 

 

N

T , P
Gibbs Free Energy and Chemical Potential
Combining
U  T S  PV  N
with G  U  T S  PV

G  N
- this gives us a new interpretation of the chemical potential: at least for the systems
with only one type of particles, the chemical potential is just the Gibbs free energy
per particle.
The chemical potential
 G 

 N T , P
 
If we add one particle to a system, holding T and P fixed, the Gibbs free energy of
the system will increase by . By adding more particles, we do not change the value
of  since we do not change the density:   (N).
Note that U, H, and F, whose differentials also have the term dN, depend on N nonlinearly, because in the processes with the independent variables (S,V,N), (S,P,N),
and (V,T,N),  = (N) might vary with N.
Example:
Sketch a qualitatively accurate graph of G vs. T for a pure substance as it changes
from solid to liquid to gas at fixed pressure.
 G 

  S
 T  P , N
- the slope of the graph G(T ) at fixed P should be –S.
Thus, the slope is always negative, and becomes
steeper as T and S increases. When a substance
undergoes a phase transformation, its entropy increases
abruptly, so the slope of G(T ) is discontinuous at the
transition.
G
solid
liquid
gas
T
S
 G 

  S
 T  P
G  ST
- these equations allow computing Gibbs
free energies at “non-standard” T (if G is
tabulated at a “standard” T)
solid
liquid
gas
T
The Minimum Free Energy Principle (P,T = const)
The total energy of the combined system (=the system of interest+the reservoir) is
U = UR+Us, this energy is to be shared between the reservoir and the system (we
assume that P and N for all the systems are fixed). Sharing is controlled by the
maximum entropy principle:
S R s U R ,U s   S R U  U s   S s U s   max
dS R  s U ,U s   dS s 
SR+s
Gs
reservoir
+system
Us
system
stable
equilibrium
Us
dU s P
dG
1
 dVs   dU s  TdS s  PdVs    s
T
V
T
T
Thus, we can enforce the maximum entropy principle by
simply minimizing the Gibbs free energy of the system
without having to know anything about the reservoir
except that it maintains a fixed T! Under these conditions
(fixed P, V, and N), the maximum entropy principle of
an isolated system is transformed into a minimum
Gibbs free energy principle for a system in the
thermal contact + mechanical equilibrium with the
reservoir.
dG T , P , N  0
Thus, if a system, whose parameters T,P, and N are fixed, is in thermal contact
with a heat reservoir, the stable equilibrium is characterized by the condition:
G  min
G/T is the net entropy cost that the reservoir pays for allowing the system to have volume
V and energy U, which is why minimizing it maximizes the total entropy of the whole
combined system.
Processes at P = const and T = const
Let’s consider the processes at P = const and T = const in general, including the
processes with “other” work:
Then
 W   PdV   Wother
dG  d U  T S  PV T , P  Q  PdV   Wother T , P  TdS  PdV
  Q T , P   Wother T , P  TdS   Wother T , P
The “other” work performed on a system at T = const and P = const in a reversible
process is equal to the change in the Gibbs free energy of the system.
In other words, the Gibbs free energy gives all the reversible work except the PV work.
If the mechanical work is the only kind of work performed by a system, the Gibbs free
energy is conserved: dG = 0.
Gibbs Free Energy and the Spontaneity of Chemical Reactions
The Gibbs free energy is particularly useful when we consider the chemical
reactions at constant P and T, but the volume changes as the reaction proceeds.
G associated with a chemical reaction is a useful indicator of weather the reaction
will proceed spontaneously. Since the change in G is equal to the maximum
“useful” work which can be accomplished by the reaction, then a negative G
indicates that the reaction can happen spontaneously. On the other hand, if G
is positive, we need to supply the minimum “other” work  Wother= G to make the
reaction go.
Helmholtz and Gibbs free energy
dU S ,V , N   T dS  PdV  dN
Potential
Variables
U (S,V,N)
S, V, N
dH S , P, N   T dS  VdP  dN
H (S,P,N)
S, P, N
F (T,V,N)
V, T, N
dF T ,V , N   S dT  PdV  dN
G (T,P,N)
P, T, N
dGT , P, N   S dT  VdP  dN
The Partition Function
For the absolute values of probability (rather
than the ratio of probabilities), we need an
explicit formula for the normalizing factor 1/Z:

1
k BT
P i  
   1
1
exp   i   exp  i 
Z
 k BT  Z
- we will often use this notation
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:
1   P i   
i
i
1
exp  i 
Z
or
Z T ,V , N    exp  i 
i
The Zustandsumme
in German
The partition function Z is called “function” because it depends on T, the spectrum (thus, V), etc.
Example: a single particle,
continuous spectrum.
(kBT1)-1
T1
T2
(kBT2)-1
0


  
d  k BT  exp  x dx  k BT
Z T    exp  
 k BT 
0
0
  

exp  
 k BT 
P  
k BT
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.

Partition Function and Helmholtz Free Energy
Now we can relate F to Z:
U  F  TS  F  T
Comparing this with the expression for
the average energy:
F
 F 1 F 
2  F 


 T 2 2 
  k BT
T
T T 
T  k BT 
T
 F 
  exp   F 
Z  exp  
 k BT 
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:
F 
  ln Z 
  k B ln Z  k BT 

S  
  T V , N
  T V , N
F 
  ln Z 
  k BT 

P  
  V T , N
  V T , N
F 
  ln Z 



  
 k BT 

  N T ,V
  N T ,V
Partition Function for a Hydrogen Atom
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:

r
3= -1.5 eV
2= -3.4 eV
1= -13.6 eV


 i 
 13.6eV 1  1 / ni2 
2
   2ni exp 
Z   di exp  

k
T
k BT
i
 B  i


(a) Estimate the partition function for a hydrogen atom
at T = 5800K (= 0.5 eV) by taking into account only
three lowest energy states.
- we can forget about the spin degeneracy – it is the same
for all the levels – the only factor that matters is n2
  
  
  
Z  exp   1   4  exp   2   9  exp   3   e0  4  e 20.4  9  e 24.2  1  5.5 109  2.8 1010  1
 k BT 
 k BT 
 k BT 
However, if we take into account all discrete levels, the full partition function diverges:

1 

13
.
6
1


2 


 13.6   2
n


2
  n  
  exp  
Z   n exp 
k BT


n 1
 k BT  n 1


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
dS R 
If we keep this term, then
1
dU R  PdVR 
T
 E  PV 

New Boltzmann factor  exp  
k BT 

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.
Average Values in a Canonical Ensemble
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:
x  x i    xi  i P i 
i
In particular, for a
canonical ensemble:
x i  
 i 
1


x

exp
 i i   k T 
Z T ,V , N  i
 B 
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):
U    i P i     i
i
i
1
1
exp  i  
Z
Z
The average values are additive. The average total
energy Utot of N identical systems is:
Another useful representation for
the average energy:
U 

i
i
i
exp  i 
i
U tot  N U
    1


exp   i   
   

 k BT   k BT

1 Z


ln Z 

ln Z  k BT 2
Z 

T


 exp        Z
i
i
thus, if we know Z=Z(T,V,N),
we know the average energy!
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
 i 

P i   di exp  
 k BT 
where di is the degree of degeneracy.
Taking the degeneracy of energy levels into
account, the partition function should be
modified:
 i 

Z   di exp  
i
 k BT 
Degenerate Energy Levels
Example: The energy levels of an electron in the
hydrogen atom:

i
2
1
r
di =2ni2
d2 =8
d1 =2
(for a continuous
spectrum, we need
another approach)
 ni   
13.6 eV
2
ni
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:
1  2ni  1  1
d i  2  2li  1  ni
 2ni2
2
lmin  0
l max  ni 1
Boltzmann Statistics: classical (low-density) limit
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). 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:
strong exchange interaction
the de Broglie wavelength
h
 ~
p
weak exchange interaction
 
n1/ 3
the average distance btw particles
h
(for N2 molecule,  ~ 10-11 m at RT)
k BT m
1/ 3
Boltzmann
h n 
 1 statistics
k BT m
applies
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.
Problem (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?
  
 2 
  exp  
  1  0.406  0.018  1.424
Z1  1  3 exp  
 k BT 
 k BT 
the average energy of a
single particle:

    i P i  
i
the same
result you’d
get from this:
 2 
  2  exp  

 k BT 
 k BT   1 eV  0.406  0.036  0.310 eV
1.424
  
 2 
  exp  

1  3 exp  
 k BT 
 k BT 
  3 exp  
 
 
1 Z
3  exp        exp  2   2 

Z 
1  3 exp     exp  2 
the average energy of N = 10 such particles:
U10  N U  10  0.310 eV  3.1 eV
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)
Z   di exp  i   1  e    e 2 
i
(b)
(c)
(d)

1 Z
  e     2 e 2 
e    2e 2 
 


Z 
1  e    e 2 
1  e    e 2 
e 2 
1  2
1
P


1  e   e 2  1  1    1  2 3
e    2e 2 
1 2
 



 
 2 
1 e  e
111
all 3 levels are populated with
the same probability
Problem 1 (cont.)
(e)
(f)
1
2
2  ln 1.1 T 
1.1
k B ln 1.1
d 
d  d
dU
CV 
N
N
dT
dT
d dT
exp  2  



1    e     2 2e  2 
e    2e  2    e     2 e  2 

 N  

2 
 
 2 
 
 2  2
k
T
1

e

e

1

e

e
 B 



 


 N 2  e    4e  2  1  e    e  2   e    2e  2  e    2e  2 

 
2
2 
1  e    e  2 
 k BT 





 e    4e  2   e  2   4e 3   e 3   4e  4   e  2   4e 3   4e  4 

2

1  e    e  2 
N 2 e    4e  2   e 3

CV
k BT 2 1  e    e  2  2
N 2

k BT 2


Low T (>>):
high T (<<):





 
2 
3 

N e  4e
e
N
k BT

e
k BT 2 1  e    e 2  2
k BT 2
N 2 e    4e 2   e 3 2 N 2
CV 

2
2



2

k BT
3 k BT 2
1 e  e
CV 

2

2




T

Problem 1 (partition function, average energy)
The neutral carbon atom has a 9-fold degenerate ground level and a 5-fold
degenerate excited level at an energy 0.82 eV above the ground level. Spectroscopic
measurements of a certain star show that 10% of the neutral carbon atoms are in the
excited level, and that the population of higher levels is negligible. Assuming thermal
equilibrium, find the temperature.
Z   di exp  i   9  5e  
i
5e  
1
P

 0.1
 

9  5e
1.8e  1
e   5 T 

k B ln 5
 5,900 K
Problem (the average values)
H
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
# of molecules within dh:
 mgh 
(area)dh
dN h  h  dh  n0 exp  
k
T
 B 
0
 mgh 
 
(area )dh
mgh

n
exp
0
0
k
T
 mgh
B


U 



H
k
T
 mgh 
 B
0 n0 exp   k BT (area )dh
mgH
k BT
H

y 

k BT




k
T
y

exp

y

dy
 B
mg
0
mgH
k BT

exp  y 
0
A
 y  exp  y dy

0
 y  exp  y 
A
0
d
 y  exp  y   exp  y   y  exp  y 
dy
k BT
dy
mg
mgH
k BT
 k BT 
 y  exp  y dy
0
mgH
k BT
0
A
 dy
0
A
A
 y  exp  y dy   A  exp  A  1  exp  A
0
0
0
  exp  y dy   y  exp  y dy
A
A
 exp  y dy   exp  A  1
0
 exp  y dy
For a very tall container (mgH/kBT ):
U  kBT
Example: energy and heat capacity of a two-level system
the slope ~ T
The partition
function:
Ei
 2= 
 0 
  
  
  exp  
  1  exp  

Z  exp  
 k BT 
 k BT 
 k BT 
The average energy:
 0 
  
1 


 
U 
   exp  
0  exp  

Z T  
 k BT 
 k BT 
 1= 0
- lnni
U
 /2
(check that the same result follows from

  

1  exp 
k
T
 B 
U 
1 Z
)
Z 
CV
0
The heat capacity at constant volume:
T
 k B  / k BT 2 exp  / k BT 
 

 U 
CV  
 





T

T
1

exp

/
k
T
1  exp  / kBT 2

V
B


Problem
Problem. 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):
  
   mgz 

exp   2  exp   1
k
T
k
T
P 2 
B
 B 

  exp   mgz 

 k T
P 1 
 1 
 1 
 B 


exp  
exp  
 k BT 
 k BT 
 mgz 

k
T
 B 
 z   0 exp  
Home work:
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:
h
k BT ln n2 / n1 
m2  m1 g
Problem
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?
 mn c 2 

exp  
 2.3 10 30 kg  3 108 m/s
k BT 
 mc2 
Pn 

  exp  

 exp  
2
 1.38 10 23 J/K 1011 K
P p 
k BT 
 mpc 



exp  

 k BT 


2

  0.86


Problem (Boltzmann distribution)
A solid is placed in an external magnetic field B = 3 T. The solid contains weakly
interacting paramagnetic atoms of spin ½ so that the energy of each atom is ±
 B,  =9.3·10-23 J/T.
(a) Below what temperature must one cool the solid so that more than 75 percent
of the atoms are polarized with their spins parallel to the external magnetic
field?
(b) An absorption of the radio-frequency electromagnetic waves can induce
transitions between these two energy levels if the frequency f satisfies he
condition h f = 2  B. The power absorbed is proportional to the difference in
the number of atoms in these two energy states. Assume that the solid is in
thermal equilibrium at  B << kBT. How does the absorbed power depend on
the temperature?
(a)
(b)
   
 2B 
P1 

 exp   1 2   exp  
P 2 
k
T
k
T
B


 B 
 2B 
  0.333
exp  
k
T
 B 
T
2B
 36.8 K
k B ln 3
The absorbed power is proportional to the difference in the number of atoms in
these two energy states:
 2B 
 2B  2B
  1  1 
 
Power  P1  - P 2   1  exp  
k
T
k
T
B
 B 

 k BT
The absorbed power is inversely proportional to the temperature.
Problem (Boltzmann distribution)
Consider an isothermic atmosphere at T=300K in a uniform gravitational field. Find
the ratio of the number of molecules in two layers: one is 10 cm thick at the earth’s
surface, and another one is 1 km thick at a height of 100 km. The mass of an air
molecule m= 5·10-26 kg, the acceleration of free fall g=10 m/s2.
 mgh 
k T
dh  (area )n0 B
N i  (area )  n0 exp  
mg
 k BT 
h1
h2
(area )n0
N10cm
N1km
k BT
mg
 mgh   mgh 
 
d 
 
exp

k
T
k
T
 B   B 
mgh1 / k BT
mgh2 / k BT
  mgh1 
 mgh2 


 

exp


exp
 

k
T
k
T
B
B



 
 mg  0m 
 mg  0.1m 
  exp  

exp  
k
T
k
T
B
B






 mg 1 105 m 
 mg 1.01 105 m 
  exp  

exp  
k
T
k
T
B
B




 5 10  26 kg 10m / s 2  0.1m 

exp  0  exp  
 23
1
.
38

10
J
/
K

300
K



 5 10  26 kg 10m / s 2 1 105 m 
 5 10  26 kg 10m / s 2 1.01 105 m 
  exp  

exp  
 23
 23
1
.
38

10
J
/
K

300
K
1
.
38

10
J
/
K

300
K




1.2 10 5
 18.5 - more air in the 10-cm-thick layer at the earth’s surface
5.69 10 6  5.04 10 6
Next … Maxwell-Boltzmann Distribution
Bose, Einstein distribution, BB