Basics of Thermal Physics
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Basics of Thermal Physics
Thermal physics is reconstructed here from information theory using two physical
ingredients-the k in entropy is now Boltzmann’s constant and the average energy
of a system has a corresponding Lagrange multiplier, , equal to 1/kT. In this
chapter we will develop the fundamentals of thermodynamics from statistical
mechanics.
Properties of Z
Various calculations can be performed relatively easily once the partition function
Z is known. Remember that the discrete partition function is defined by the expression
Z   e  Ei .Two very useful relations are:
i
E 

ln Z

(1)
and
2
  2 ln Z ,

where  is the standard deviation of energy.
2
(2)
1.
Use the definition of Z to derive Eqs.(1) and (2).
2.
Given that an electron in a magnetic field B has only the energies E   B, find
the partition function and use it to calculate the average energy. ans:
Thermodynamic Entropy
We want to demonstrate (again) that, with sufficient qualifications, information
entropy is mathematically equivalent to thermodynamic entropy. That is, for reversible
systems,
dQ
dS 
T
Consider the differential of the average energy condition E   p j E j such that
dE   E j dp j  p j dE j
This can be compared with the first law of thermodynamics,
dE  dQ  dW
(3)
(4)
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where dQ and dW represent heat going into the system and work done on the system. We
identify dQ as the energy change due to the system occupying different available energy
levels. Similarly, we identify dW with a mechanical shift in the energy levels.
A system that undergoes a reversible process is able to return to its original
configuration with no level changes. In such cases, dE  dQ . Since the partition
function is central to calculations in statistical mechanics, it is useful to express entropy
(as well as energy) in terms of Z:


S  k  pi ln pi k E  ln Z .
(5)
i
Taking the differential of S for a reversible process gives the desired equivalence,



dQ
dS  k dE  Ed 
ln Z   kdE 
,

T


where we used Eq.(1) to affect the simplification.
3.


Derive S  k  E  ln Z using S  k  pi ln pi and the canonical distribution.
i
Classical thermodynamics defines entropy with the relation
dQrev
dS 
T
It is worth repeating that, in our dynamic interpretation of temperature, entropy is a
measure of the average number of state changes accessible to the system. Note that this
applies only to reversible increments of heat.
Heat added irreversibly cannot be fully utilized for state transitions. Consequently,
the dS must be greater for an orderly reversible process than for an irreversible process .
An important conclusion is
dQrev dQ
(6)
dS 

,
T
T
where the equality applies only for reversible processes (and closed systems).
Reversibility is a convenient fiction because entropy is a state variable*—a real,
irreversible process can be replaced by a reversible process that brings the system to the
same end point.
*
A state variable is one that does not depend on its history. For example, the gravitational potential energy
of an aircraft sitting on an airfield does not depend on how high it once flew.
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Outline of Thermodynamics I-Basics
Equations (4) and (6) correspond to the first and second laws of thermodynamics:
dE  dQ  dW
(4)
dS 
dQ
T
(6)
Combining these for reversible processes we have
(7)
dE  T dS  dW .
This is the fundamental expression of thermodynamics for reversible systems (and we
almost always “replace” real systems with reversible equivalents). Often, work is done on
the surroundings by pressure P altering volume V so that dW  PdV . We will see other
forms of work in subsequent chapters.
We usually know some relation between thermodynamic variables. Some
examples include state equations like the ideal gas relation, energy relations like
E  V T 4 , or entropy relations. Frequently the problem then is to derive other relations
between variables.
Since we know in principle how to find entropy, we illustrate the procedure for
relating variables to entropy. Write a physical expression for dS based on Eq.(7) and
compare this with a mathematical identity:
dE PdV
Physics dS 

T
T
 S 
 S 
Math
dS    dE  
 dV
 E  V
 V  E
The comparison gives
1
P
 S 
 S 
(8)
and   
  
 E  V T
 V  E T
This illustrates a pattern whereby various alterations of Eq.( 7) –the physics– is compared
to the corresponding mathematical identity. This enables thermodynamicists to wring
varied relations from some known quantity. In the illustration, we assumed S was known.
4.
Later we will find S for an ideal gas has the form S  kN ln V  f (T , E ) . Use this
to find an expression for P in terms of V and T. [ans. PV = nRT]
5.
Starting with the differential dU  T dS  PdV (physics) write the corresponding
 U 
 U 
mathematical identity and show 
  T and 
  P .
 S V
 V  S
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Outline of Thermodynamics II-Potentials
Some analyses or problems are best expressed in terms of a particular set of
independent variables. For instance, a process that is both adiabatic (dQ = 0 or
S = constant) and isochoric (V = constant) is conveniently expressed by a function of S
and V, E  E  S ,V  . Similarly, an isothermal and isochoric process is easily described by
a function of T and V, F  F T ,V  . A number of “thermodynamic potentials” like these
are available.
The following are the most familiar thermodynamic potentials:
E(S,V) System Energy
dE  TdS  PdV
(9)
S(E,V) System Entropy
dS 
dE PdV

T
T
F(T,V) Helmholtz Free Energy
F  E  TS
(11)
G  F  PV
(12)
G(T,P) Gibbs Free Energy
6.
(10)
Demonstrate that Helmholtz Free Energy, F  E  TS , is a function of
independent variables T and V. (That is, show that dF is an expression that varies
with dT and dV.)
Helmholtz Free Energy F is very useful for converting statistical mechanical
information into thermodynamic expressions. In particular, we can show that F is simply
related to the partition function by the now familiar expression
F   kT ln Z
(13)
Once F is known, P and S can be found from the relations
 F 
 F 
(14)
P    and S   
 V  T
 T  V
7.
Derive Eqs.(14).
One of the most usual routines for applying statistical mechanics is (i) calculate
the partition function, (ii) find F from (13), and (iii) generate state equations from
Eqs.(14).
Basics of ThermalPhysics
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In one dimension an increment of work is the product of an average force f and a
distance dL The combined first and second law for reversible processes is then
dE  T dS  f dL
where, in this case, f is the force exerted by the system on the environment (the
one-dimensional equivalent of pressure).
(a) Show that for this system,
 F 
 F 
f  
 and S  

 L T
 T  L
(b) Show that the partition function for a (fictitious) one-dimensional ideal gas of
N particles in a “box” of length L is
N
 2 m  2 L 

Z  

   h 
(c) Derive the state equation for this one-dimensional gas,
f L  NkT
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