1. Chapter 1: Statistical Basis of Thermodynamics

2. Macroscopic & Microscopic States
Start with a macroscopic system characterized by
N identical particles confined to volume V.
N 1024
We often will be considering the
Thermodynamic Limit
which is defined as
N   & V 
These limits are taken so that the Particle Density
n = (N/V) = constant.

3. E  i ni i, where N  i ni . i  energy of particles of type i
If the N particle system is such that there are no interactions between particles (unrealistic!) the total energy E of the system can be written:
E  i ni i, where
i  energy of particles of type i
ni  number of particles of type i .
Of course, it also must be true that
N  i ni .
If we model the microscopic physics of the system with Quantum Mechanics, the single particle energies i are discrete. If we use classical mechanics, they are continuous.

4. The number of particles N,
For such a system, if we take specific values for
The number of particles N,
The volume V
& The total energy E,
we are said to have specified the system
Macrostate.

5. In an external magnetic field B
We are implicitly assuming 1 external parameter, the volume V. So it may seem like we are only talking about gases. But, most of what we say will be valid for systems with other external parameters.
For example, consider a system containing
N magnetic moments 
In an external magnetic field B
& with Total energy E.
The system Macrostate is specified by stating N, B & E.
B is an external parameter analogous to V

6. In contrast to the system Macrostate, as discussed in the Review of Undergrad Statistical Physics that we did, the system
Microstate
in the Quantum picture is given by specifying every quantum number of every particle of the system!
In the Classical picture it is given by stating the system “location” in an abstract 6N dimensional phase space.

7. The Fundamental Hypothesis: Each allowed microstate is
As we also discussed in the Review of Undergrad Stat. Physics:
There are a HUGE number of Microstates that can be associated with a given Macrostate.
Then, to treat the system with probability, we must use
The Fundamental Hypothesis:
Each allowed microstate is
equally probable.

8. (N,V,E)  For such a system, lets define the function
the number of allowed microstates for a given macrostate.
In our review we showed some results for the functional form of (N,V,E).
However, for the following, lets assume that we know its functional form.

9. From our review, we know that
(N,V,E)
is the primary connection between the macroscopic physics & the microscopic physics of our system!

10. Significance of (N,V,E)
Consider an isolated system A0. It is divided into two sub-systems A1 & A2.
1(N1,V1,E1)  Number of accessible microstates of A1
2(N2,V2,E2)  Number of accessible microstates for A2
Assume that A1 & A2 were initially separately in equilibrium. The isolated system A0 is formed when A1 & A2 are brought together & allowed to interact.
At first, they exchange energy
but not particles.

11. The situation is shown schematically in the figure:
Because it is isolated, the combined system A0 has a constant energy E0:
E0 = E1 + E2 = constant

12. E0 = E1 + E2
= constant
N1 & N2 fixed
Using The Fundamental Postulate, A1 is equally likely to be in any one of its allowed microstates 1(E1) & A2 is equally likely to be in any one of its allowed microstates 2(E2). (Focusing on the E dependences, suppressing the V dependences.)
Further, the combined system A0 is equally likely to be in any one its allowed microstates
1(E1) 2(E2) = 1(E1) 2(E0 – E1)
 0(E0, E1)

13. Let E1  Most probable energy of system A1
An outline of a calculation
of E1 follows
E2 =
E - E1 E1
E = E1 + E2
The probability finding of A1 with energy E1 is
proportional to the product of the number of
accessible states of A1 times the number of
accessible states of A2 . This must be Consistent
with Energy Conservation: E = E1 + E2.

14. Each configuration is equally probable,
E = E1 + E2 E1
E2 =
E - E1
Stated another way:
Each configuration is equally probable,
but, the number of states having a particular
energy E1 is huge & unknown. Find E1 by finding
the energy where the left side of the above equation
has a maximum. Note: Since entropy S = kBln,
maximizing  is the same as maximizing the entropy

15. Some Math Details Combine (1) & (2): Start with (1): Take the natural
log of both sides:
Find E1 where this is
max by setting the
derivative = (2):
Combine (1) & (2):

16. Definition of the parameter β
Combine (1)
& (2):
Energy is
Conserved!  dE1 = -dE2
This is the Thermal Equilibrium
Condition!!
Definition of the parameter β

17. A Fundamental Statistical Definition! A System’s Temperature!
So, at Equilibrium, it must be true that:
The Thermal Equilibrium
Condition!!
Definition of β
A Fundamental Statistical Definition!
The physical interpretation of β is that it is a measure of
A System’s Temperature!

18. In thermodynamics, we DEFINE the Kelvin Temperature Scale such that:
Also, DEFINE the Entropy S of a system as:
(kB=  J/K = Boltzmann’s constant)
In thermodynamics, we DEFINE the
Kelvin Temperature Scale such that:
But we just defined:
So,   [1/(kBT)]

19. A “Statistical” Definition of Entropy:
 Entropy & Temperature are both related to the number of accessible states Ω(E).
A “Statistical” Definition of Entropy:
A “Statistical” Definition of Temperature:

20. (Obvious, but we proved it!)
 The Fundamental Postulate leads to:
1. An Equilibrium Condition: Two systems
are in thermal equilibrium when their
temperatures are equal.
(Obvious, but we proved it!)
2. Maximum Entropy for the coupled
systems when they are at equilibrium. As
we know, this is related to the
The 2nd Law of Thermodynamics.
3. The 3rd Law of Thermodynamics also.

21. Assume Thermal Interactions Only
Consider again the 2 macroscopic systems A1 & A2, interacting & in equilibrium. The combined system A0 = A1 + A2, is isolated.
Assume Thermal Interactions Only
Recall that in a pure
Thermal Interaction,
no mechanical work is
done. So the energy
exchange between
A1 & A2, is a pure A2 A1
Heat Exchange!!

22. P(E1)  [2π (ΔE)2]-½exp[-(E1 -Ē1)2/(2(ΔE)2)]
As we said, the probability P(E1) of finding the
combined system A0 = A1 + A2 in a state for which
the system A1 has energy E1 is obtained by finding
the energy E1 at which the following function is
maximized. P(E1) is proportional to:
After a tedious derivation, it can be shown that an
approximate form for P(E1) is a Gaussian:
P(E1)  [2π (ΔE)2]-½exp[-(E1 -Ē1)2/(2(ΔE)2)]
It can be shown that the width has the form
 = ΔE  (Ē1)/(f)½ .
Note: if f = 1024, (f)½ = 1012 so ΔE  (Ē1)/(f)½  (10-12)Ē1 .
So the Gaussian with is, very very tiny!

23. For f = 1024, (f)½ = 1012, so ΔE  (Ē1)/(f)½  (10-12)Ē1
In other words, the probability P(E1) of finding the
combined system A0 = A1 + A2 in a state for
which the system A1 has an energy E1 looks like:
P(E1)
ΔE  (10-12)Ē1
So,
P(E1)  Dirac
delta function
centered at Ē1!
ΔE  (Ē1)/(f)½ E1 Ē1
For f = 1024, (f)½ = 1012, so ΔE  (Ē1)/(f)½  (10-12)Ē1

24. Mechanical Interaction, They Do Work on each other!
Consider 2 macroscopic systems A1 & A2 interacting & in equilibrium. The combined system A0 = A1 + A2 , is isolated.
Assume only
Mechanical Interactions
Recall that in a pure
Mechanical Interaction,
mechanical work is done.
So, the energy exchange between A1 & A2
occurs because A2 A1
They Do Work on each other!

25. X  -(E/x) With Mechanical Interactions A1 & A2
interact by Doing Work on each other
Recall also that doing
mechanical work means
that at least one external
parameter (x) of A1 changes. A2 A1
Further, this change in x is characterized by
a generalized force
X  -(E/x)

26. Look for V1  Most probable
Now consider the volume
dependences of the number of
accessible states  in a similar
way to our earlier consideration
of their energy dependences.
V2 =
V - V1 V1
V = V1 + V2
Look for V1  Most probable
volume of system A1
Find V1 by finding the volume where the left side
of the following equation has a Maximum:
(V1, V – V1) = 1(V1) 2(V – V1)

27. [∂ln(1)/∂V1] = [∂ln(2)/∂V2]
This leads to the following condition for
mechanical equilibrium between A1 & A2:
[∂ln(1)/∂V1] = [∂ln(2)/∂V2]
After more manipulation, it is shown that the
physical interpretation of the quantity
[∂ln()/∂V] is that it is related to the mean
pressure p as
p  kBT[∂ln()/∂V]
More generally, the mean generalized force
X has the form: X  kBT[∂ln()/∂x]

28. [∂ln(1)/∂V1] = [∂ln(2)/∂V2] at equilibrium, the pressures
The condition for mechanical
equilibrium between A1 & A2
[∂ln(1)/∂V1] = [∂ln(2)/∂V2]
therefore tells us (no surprise!) that,
at equilibrium, the pressures
are equal!
p1 = p2

29. For f = 1024, (f)½ = 1012, so ΔX  <X1>/(f)½  (10-12)<X1>
In general, for an external parameter x & generalized force X,
much tedious derivation shows that the probability P(X1) of
finding the combined system A0 = A1 + A2 in a state for
which the system A1 has generalized force X1 looks like:
P(X1)
ΔX  <X1>/(f)½
ΔX  (10-12)<X1> X1
<X1>
For f = 1024, (f)½ = 1012, so ΔX  <X1>/(f)½  (10-12)<X1>

30. For f = 1024, (f)½ = 1012, so Δp  <p1>/(f)½  (10-12)<p1>
If the external parameter for a gas is the volume, x = V, then we’ve
seen that the generalized force is the pressure p, X = p. So the
probability P(p1) of finding the combined system A0 = A1 + A2
in a state for which the system A1 has pressure p1 looks like:
P(p1)
Δp  <p1>/(f)½
Δp  (10-12)<p1> p1
<p1>
For f = 1024, (f)½ = 1012, so Δp  <p1>/(f)½  (10-12)<p1>

31. Both Thermal & Mechanical Interactions Occur at the Same Time
Consider 2 macroscopic systems A1, A2, interacting & in equilibrium. The combined system A0 = A1 + A2 is isolated. Assume that
Both Thermal & Mechanical Interactions Occur at the Same Time A2 A1

32. For f = 1024, (f)½ = 1012, so ΔE  Ē1/(f)½  (10-12)Ē1
The probability of finding the combined system
A0 = A1 + A2 being in a state for which the system
A1 has energy E1 is as before. So,
P(E1)
ΔE  (10-12)Ē1
ΔE  Ē1/(f)½ E1 Ē1
For f = 1024, (f)½ = 1012, so ΔE  Ē1/(f)½  (10-12)Ē1

33. For f = 1024, (f)½ = 1012, so ΔX  <X1>/(f)½  (10-12)<X1>
The probability of finding the combined system
A0 = A1 + A2 being in a state for which the system
A1 has a mean generalized force X1 is as before, so
P(X1)
ΔX  <X1>/(f)½
ΔX  (10-12)<X1> X1
<X1>
For f = 1024, (f)½ = 1012, so ΔX  <X1>/(f)½  (10-12)<X1>

34. Now, lets relax that assumption & allow them to exchange particles.
So far, we’ve assumed
that the 2 macroscopic systems A1, A2, can exchange energy (by doing work W on each A2 A1
other & exchanging heat Q), but that they cannot exchange particles.
Now, lets relax that assumption & allow them to exchange particles.
Follow the same type of procedure we used for thermal & mechanical interactions to

35. Now Particle Exchange is Allowed.
So, consider 2 macroscopic systems A1 & A2 interacting & in equilibrium. The combined system A0 = A1 + A2 , is isolated. A2 A1
Now Particle Exchange is Allowed.
So, consider the Energy Exchange between
A1 & A2 due to the fact that particles are
moving across the boundary between them.
To discuss this, we first need to discuss the
concept of the Chemical Potential.

36. Chemical Potential
The Chemical Potential  of an N particle system is defined as
the energy required to remove one particle from the system & move it out to an infinite distance away.

37. Look for N1  Most probable
Consider the dependences of the
number of accessible states  on
the particle number N in a similar
way to our earlier considerations
of the E & V dependences.
N2 =
N - N1 N1
N = N1 + N2
Look for N1  Most probable
particle number for system A1
Find N1 by finding the particle number where the left
side of the following equation has a Maximum:
(N1, N – N1) = 1(N1) 2(N – N1)

38. [∂ln(1)/∂N1] = [∂ln(2)/∂N2]
This leads to the following condition for
Chemical equilibrium between A1 & A2:
[∂ln(1)/∂N1] = [∂ln(2)/∂N2]
After more manipulation, it is shown
that the physical interpretation of the
quantity [∂ln()/∂N] is that it is related
to the mean Chemical Potential  as
  kBT[∂ln()/∂N]

39. [∂ln(1)/∂N1] = [∂ln(2)/∂N2] at equilibrium, the Chemcial
The condition for Chemical
equilibrium between A1 & A2
[∂ln(1)/∂N1] = [∂ln(2)/∂N2]
therefore tells us (no surprise!) that,
at equilibrium, the Chemcial
Potentials are equal!
1 = 2