Chapter 3: Interactions and Implications.
Start with Thermodynamic Identities
S U ,V , N   kB ln  U ,V , N 
 S 
 S 
 S 
dS  
 dU  
 dV  
 dN
 U  N ,V
 V  N ,U
 N U ,V
1
 S 

 
 U V , N T
P
 S 



 V U , N T
 S 

 ?
 N U ,V
Diffusive Equilibrium
and Chemical Potential
UA, VA, NA UB, VB, NB
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 sub-systems, the
molecules in A and B are the same ).
For sub-systems in
diffusive equilibrium:
 S AB 


0

U
A V A , N A

 S A  SB

 N A  NB
S
In equilibrium,
 S AB 


0
 N A U A , V A

 S 

 
T
 N U ,V
TA  TB
 A  B
UA
 S 

 N U ,V
  T 
- the chemical
potential
NA
Sign “-”: out of equilibrium, the system with the larger S/N will get more
particles. In other words, particles will flow from from a high /T to a low /T.
Chemical Potential: examples
Einstein solid:
consider a small one, with N = 3 and q = 3.
 ( N  3, q  3) 
q  N  1 !  10


dS 
Thus, for this system
Monatomic ideal gas:
S ( N  3, q  3)  kB ln10
q !( N  1) !
let’s add one more oscillator:
U
 
  
  N S
 U 

  N V , S
  
S ( N  4, q  3)  kB ln 20
1

dU  d N
T
T
To keep dS = 0, we need to decrease the
energy, by subtracting one energy quantum.
 U 
  
  
  N S
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
 V  2 m

VT 3 / 2 
 S 
 
  T 
   k BT ln   2 k BT     k BT ln  g m

N 
 N U ,V
 
 N  h

At normal T and P, ln(...) > 0, and  < 0 (e.g., for He,  ~ - 5·10-20 J ~ - 0.3 eV.
Sign “-”: usually, by adding particles to the system, we increase its entropy.
To keep dS = 0, we need to subtract some energy, thus U is negative.
The Quantum Concentration
3/ 2
3/ 2
2


 V  2 m

 h3


h
P 


   k BT ln
   k BT ln   2 k BT    k BT ln n

3/ 2
5/ 2 

kBT  
 
  2  m kBT  
 N  h
 2m

n=N/V – the concentration of molecules
0
The chemical potential increases with the density of the gas or with
its pressure. Thus, the molecules will flow from regions of high
density to regions of lower density or from regions of high pressure
to those of low pressure .
3/ 2
2
 

 n 

h


  k BT ln 
  k BT ln n

n 
  2  m kBT  
 Q
 2 m

nQ   2 kBT 
 h

dB
3/ 2

when n
increases
when n  nQ,   0
- the so-called quantum concentration (one particle per
cube of side equal to the thermal de Broglie wavelength).
When nQ >> n, the gas is in the classical regime.
h
h
 
p
m kBT
nQ 
1
dB 3
 m kBT 


2
 h 
3/ 2
At T=300K, P=105 Pa , n << nQ. When n  nQ, the quantum statistics comes into play.
Entropy Change for Different Processes
The partial derivatives of S play very important roles because they determine
how much the entropy is affected when U, V and N change:
Type of
interaction
Exchanged
quantity
Governing
variable
thermal
energy
temperature
1  S 


T  U V , N
volume
pressure
P  S 


T  V U , N
particles
chemical
potential
mechanical
diffusive
Formula

 S 
 

T

N

U ,V
The last column provides the connection between statistical physics and
thermodynamics.
Thermodynamic Identity for dU(S,V,N)
S  S U ,V , N 

if monotonic as a function of U (“quadratic”
degrees of freedom!), may be inverted to give
U  U S,V , N 
 U 
 U 
 U 
dU  
dS

dV





 dN
 S V , N
 V  S , N
 N  S ,V
 U 

  T

S

 N ,V
compare with
 U 

   P

V

 N ,S
 S 
1

 
  U  N ,V T
d U  T d S  P dV   d N
pressure
 U 

  

N

V , S
chemical potential

shows how much the system’s energy
changes when one particle is added to the
system at fixed S and V. The chemical
potential units – J.
- the so-called thermodynamic
identity for U
This holds for quasi-static processes (T, P,  are well-define throughout the system).
Thermodynamic Identities
With these abbreviations:
- the so-called
thermodynamic identity
dU  TdS  PdV  dN
 shows how much the system’s energy changes when one particle is added to the
system at fixed S and V. The chemical potential units – J.
 is an intensive variable, independent of the size of the system (like P, T, density).
Extensive variables (U, N, S, V ...) have a magnitude proportional to the size of the
system. If two identical systems are combined into one, each extensive variable is
doubled in value.
The thermodynamic identity holds for the quasi-static processes (T, P,  are welldefine throughout the system)
The 1st Law for quasi-static processes (N = const):
 S 
 S 
 S 
dS  
dU

dV





 dN
 U  N ,V
 V  N ,U
 N U ,V
dU  TdS  PdV
1
P

dS  dU  dV  dN
T
T
T
This identity holds for small changes S provided T and P are well defined.
The coefficients may
be identified as:
 S 
1

 
  U  N ,V T
S 
P



  V  N ,U T
 S 


  
T
  N V ,U
The Equation(s) of State for an Ideal Gas
Ideal gas:
(fN degrees of freedom)
 U ,V , N   g N  V NU f N / 2
S U ,V , N   Nk B ln  g  N VU f / 2 
The “energy” equation of state (U  T):
1  S 
f
1

  N kB
T  U V , N 2
U
U
f
N kB T
2
The “pressure” equation of state (P  T):
N kB
 S 
P T 
 T
V
 V U , N
PV  N k BT
- we have finally derived the equation of state of an ideal gas from first principles!
In other words, we can calculate the thermodynamic information for an isolated
system by counting all the accessible microstates as a function of N, V, and U.
Ideal Gas in a Gravitational Field
Pr. 3.37. Consider a monatomic ideal gas at a height z above sea level, so each
molecule has potential energy mgz in addition to its kinetic energy. Assuming that
the atmosphere is isothermal (not quite right), find  and re-derive the barometric
equation.
U  U kin  Nmgz
note that the U that appears in the Sackur-Tetrode
equation represents only the kinetic energy
d U  T d S  P dV   d N
 U 

 N  S , V
 
3/ 2
 V 2  m
 
  ( z )  mgz  k BT ln 
k BT    mgz

2
 
 N  z   h
In equilibrium, the chemical potentials between any two heights must be equal:
3/ 2
3/ 2
 V  2 m
 V  2 m
 
 
 k BT ln 
 2 k BT    m gz  k BT ln 
 2 k BT  
 
 
 N ( z )  h
 N (0)  h
kBT ln N ( z)  mgz  kBT ln N (0)
N ( z )  N (0)e

mgz
k BT
An example of a non-quasistatic adiabatic process
Caution: for non-quasistatic adiabatic processes, S might be non-zero!!!
P
2
S = const
along the
isentropic
line
2*
1
Vf
Vi
V
Pr. 3.32. A non-quasistatic compression. A cylinder with
air (V = 10-3 m3, T = 300K, P =105 Pa) is compressed (very
fast, non-quasistatic) by a piston (A = 0.01 m2, F = 2000N,
x = 10-3m). Calculate W, Q, U, and S.
U   Q   W
holds for all processes,
energy conservation
U  TS  PV
quasistatic, T and P are welldefined for any intermediate state
quasistatic adiabatic  isentropic
W 
Vf


 P(V )dV  PV  const  PiVi

Vi
PiVi  1
1  PiVi  Vi




  1 V f  1 Vi  1    1  V f

 1

PiVi 
x 

  1
1 
  1 
x 


 PiVi




Q = 0 for both

non-quasistatic adiabatic
Vf
1
V V  dV
i
 1

 1


x
 1 105 Pa 103 m3 10 2  1J
x
Vf
 W   P  dV  P Vi  V f 
Vi
 2 105 Pa 102 m 2 103 m2
 2J
The non-quasistatic process results
in a higher T and a greater entropy
of the final state.
Direct approach:
S U , V , N   N k B ln V 
Q0
adiabatic quasistatic  isentropic
f
N k BT
N k B T  
V
2
V
VT
f /2
f
N k B ln U  k B ln f N 
2
U   W
 const
Vf
f
2 Vf
S  N k B ln  N k B ln
0
Vi 2
f Vi
f
N k B T
2
T f  Vi

Ti  V f
  PV




2/ f
adiabatic non-quasistatic
Uf
V f  Vi f
U f  Ui
f
S  Nk B ln  Nk B ln
 Nk B
 Nk B
Vi 2
Ui
Vi
2
Ui
Vf

Pi
1
U  Pi V  PV  Pi V
V  U 

Ti
Ti
Ti
Ti
5
5
5
P  Pi  V   2  10  10   10




Ti
V
 10  2
V
300
U  W  2 J
Ui 
1
J/K
300
f
5
5
N k BTi  Pi Vi  10 5 Pa 10 -3 m 3  250 J
2
2
2
P
2
U = Q = 1J
To calculate S, we can consider any quasistatic
process that would bring the gas into the final state (S is
a state function). For example, along the red line that
coincides with the adiabata and then shoots straight up.
Let’s neglect small variations of T along this path ( U
<< U, so it won’t be a big mistake to assume T  const):
1
Vf
Vi
S  0 (adiabata) 
V
Q 2J  1J
1


J/K
T
300 K 300
The entropy is created because it is an irreversible,
non-quasistatic compression.
P
2
U = Q = 2J
For any quasi-static path from 1 to 2, we must have the
same S. Let’s take another path – along the isotherm
and then straight up:
1
isotherm: S 
T
1
Vf
Vi
V
Total gain of entropy:
Vf
Pi Vi
P
(
V
)
dV


T
Vi
dV Pi Vi  V f 
V V  T ln Vi 
i
Vf
Pi Vi x
105 P a 10-3 m 3 10-2
1



J/K
T x
300K
300
U Q
2J
2
“straight up”:
S 
 

J/K
T
T 300 K 300
1
2
1
S  
J/K
J/K 
J/K
300
300
300
P
The inverse process, sudden expansion of an ideal gas
(2 – 3) also generates entropy (adiabatic but not
quasistatic). Neither heat nor work is transferred: W = Q
= 0 (we assume the whole process occurs rapidly enough
so that no heat flows in through the walls).
2
3
1
Vf
Vi
V
Because U is unchanged, T of the ideal gas is unchanged.
The final state is identical with the state that results from a
reversible isothermal expansion with the gas in thermal
equilibrium with a reservoir. The work done on the gas in
the reversible expansion from volume Vf to Vi:
Wrev  N k BT ln
Vf
Vi
The work done on the gas is negative, the gas does positive work on the piston in an
amount equal to the heat transfer into the system
Qrev  Wrev  0
S 
Qrev  Wrev
V
PV V
1

 NkB ln i  i i

J/K
T
T
Vf
Ti Vi
300
Thus, by going 1  2  3 , we will increase the gas entropy by
S1 23 
2
J/K
300
Systems with a “Limited” Energy Spectrum
The definition of T in statistical mechanics is
consistent with our intuitive idea of the
temperature (the more energy we deliver to a system,
the higher its temperature) for many, but not all
systems.
1
 S 

T  
  U V , N
“Unlimited” Energy Spectrum
the multiplicity increase monotonically with U :   U f N/2
ideal gas in thermal equilibrium
S
 S 

 0

U

V , N
self-gravitating ideal
gas
(not in thermal
equilibrium)
S
Pr. 1.55
 2S 

 0
2 
 U V , N
S P  CP dT 
T
U
U
1  S  T


T  U V , N
T
T>0
T>0
U
C
U
C
 S 
   P
 T  P T
At T 0, the graph goes to 0 with
zero slope. At high T, the rate of
the S increase slows down (CP 
const). When solid melts, there is a
large S at T = const, another jump
–
at
liquid–gas
phase
transformation.
S
C
 U 
CV  

 T V
CV  0
Pr. 3.29. Sketch a graph
of the entropy of H20 as a
function of T at P = const,
assuming that CP is
almost const at high T.
U
U
CV  0
ice
water
vapor
T
E
“Limited” Energy Spectrum: two-level systems
e.g., a system of non-interacting spin-1/2 particles in external
magnetic field. No “quadratic” degrees of freedom (unlike in an ideal
gas, where the kinetic energies of molecules are unlimited), the
energy spectrum of the particles is confined within a finite interval
of E (just two allowed energy levels).
S
S
1  S 

 0
T  U  N ,V
0.7
0.6
S / NkB
the multiplicity and entropy
decrease for some range of U
2NB
0.5
0.4
0.3
T>0 T<0
0.2
0.1
0.0
-1.0
-0.5
0.0
U / NB
0.5
1.0
U
U
T
U
in this regime, the system is
described by a negative T
Systems with T < 0 are “hotter” than the systems at any positive
temperature - when such systems interact, the energy flows from a
system with T < 0 to the one with T > 0 .
½ Spins in Magnetic Field

B

N - the number of “down” spins
N  N  N
E
The magnetization:
M   N  N    2N  N 
E2 = + B
an arbitrary choice
of zero energy
0
E1 = - B
N - the number of “up” spins
The total energy of the system:
U  MB   B N  N    B N  2N 
 - the magnetic moment of an individual spin
Express N and N with N and U ,
N 
N
2

U 
1 

 NB 
N 
N
2

U 
1 

 NB 
Our plan: to arrive at the equation of state for a two-state paramagnet
U=U(N,T,B) using the multiplicity as our starting point.
1
  S ( N ,U ) 

  U =U (N,T,B)
T

 (N,N)  S (N,N) = kB ln  (N,N) 
 U

From Multiplicity – to S(N) and S(U)
The multiplicity of any macrostate
with a specified N:
 ( N , N ) 
N!
N ! N !


S N , N  
N!

  ln N!  ln N  !  lnN  N  !
 ln

kB
 N  ! N  N  ! 
 ln N! N ln N  N   N ln N  N  ln N   N  N  lnN  N  
N  N  N

N
U 
N
U 

N

1

,
N

1







U   B  N  2N  
2
N

B
2
N

B






N
U  N 
U  N 
U  N 
U  
 ln  1 
  1 
 ln  1 

S  N , U   k B  N ln N  1 
2
N

B
2
N

B
2
N

B
2
N

B


  


  


N
U  
U  N
U  
U 
 ln1 
  1 
 ln1 

 k B  N ln 2  1 
2
N

B
N

B
2
N

B
N

B

 


 


0.7
Max. S at N = N  (N= N/2): S=NkBln2
S / NkB
0.6
0.5
0.4
0.3
T>0 T<0
0.2
ln2  0.693
0.1
0.0
-1.0
-0.5
0.0
U / NB
0.5
1.0
1  S 
 
N
U  
U  N
U  
U 









N
ln
2

1

ln
1


1

ln
1

  kB

T  U  N , B
U 
2 
NB  
NB  2 
NB  
NB 
U 

1


 1


U 
1
1
U 
1 
kB
N

B

 
 
 kB 
ln1 

ln1 
ln

NB  2 B 2 B 
NB  2 B  2  B  1  U 
 2 B 

NB 

From S(U,N) – to T(U,N)
The same in terms of N and N :
N
1

N
2
 
N N 
1
2 
2  B   N  U /  B 

T
ln
k B   N  U /  B 
1
S
 N ln N  N ln N   N  N lnN  N 
kB
U 
U
1

N 
N 
N  B 
1
k
k
N B

  B ln      B ln   
T 2 B  N  
2 B  N  
U  1 U
N B
N  B 
Energy
 2 B 
 E  E 
 exp  
  exp  

N
k
T
k
T
B
 B 


N
Boltzmann factor!
 E  E 
 exp   

N
k BT 

N
E
E
1
2 B   N 
2  B   N  U /  B 

T
ln

 

ln

k B   N 
k B   N  U /  B 
The Temperature of a
Two-State Paramagnet
Energy
E2
T = +  and T = -  are (physically)
identical – they correspond to the same
values of thermodynamic parameters.
N
 E  E1 

 exp  2
N
k
T
B


E2
T
E1
E2
E1
E1
N B
E2
- N B
E1
Systems with neg. T are “warmer” than
the systems with pos. T: in a thermal
contact, the energy will flow from the
system with neg. T to the systems with
pos. T.
0
U
Energy
E2
E1
N
 E  E1 

 exp  2
N
k BT 

1
The Temperature of a Spin Gas
The system of spins in an external magnetic field. The internal energy in this
model is entirely potential, in contrast to the ideal gas model, where the
energy is entirely kinetic.
Boltzmann distribution
ni  e
At fixed T, the number of spins
ni of energy Ei decreases
exponentially
as
energy
increases.
 ln ni 
Ei
k BT
the slope
T
Ei
E6
Ei
k BT
E5
- lnni
E4
B
E3
spin 5/2
(six levels)
E2
E1
Ei
T=0
- lnni

Ei
T=
- lnni
B
Ei
no T
- lnni
For a two-state system, one can always introduce T - one can always fit an exponential
to two points. For a multi-state system with random population, the system is out of
equilibrium, and we cannot introduce T.
The Energy of a Two-State Paramagnet
1
  S ( N ,U ) 
 (N,N)  S (N,N) = kB ln  (N,N)  T  
  U =U (N,T,B)

U


The equation of state of a two-state paramagnet:
 N U /  B 
1
k

 B ln
T 2 B  N  U /  B 
 1  e 2  B / k BT
U  N  B 
2  B / k BT
1

e

U
U
N B
1
- N B

B
   N  B tanh

 k BT 

 B/kBT
T
- N B
U approaches the lower limit (-NB) as T decreases or, alternatively,
B increases (the effective “gap” gets bigger).
S
S(B/T) for a Two-State Paramagnet
NkBln2


S N , N  
N!
  N ln N  N  ln N   N  N  lnN  N  
 ln

kB
 N  ! N  N  ! 
T>0 T<0
Problem 3.23 Express the entropy of a two-state
paramagnet as a function of B/T .
N  N (T ) ?
B
k BT
x
U   B N  2 N 
 B N  2 N     N  B tanh x
- N B
0
B

U   N  B tanh
 k BT 
N  N
1  tanh x
2
S N , x 
 1  tanhx   1  tanhx   1  tanhx   1  tanhx 
 ln N  

 ln  N
 ln  N


N kB
2
2
2
2

 
 
 

 1  tanhx  1  tanhx   1  tanhx  1  tanhx 
 
 ln 
 ln 
 

2
2
2
2

 
 
 

cosh x  sinh x
ex
1  tanhx 

cosh x
cosh x
N B
cosh x  sinh x
e x
1  tanhx 

cosh x
cosh x
U
S(B/T) for a Two-State Paramagnet (cont.)
 ex 
 e x 
S N , x 
ex
e x



ln
ln
N kB
2 cosh x  2 cosh x  2 cosh x  2 cosh x 
ex
e x
x  ln2 cosh x  
 x  ln2 cosh x 

2 cosh x
2 cosh x
 e x  ex   e x  ex 
  
 ln2 cosh x   ln2 cosh x   x tanhx
  x
2
cosh
x
2
cosh
x

 

 

B 
B  B
B 
  N k B ln 2 cosh
 
S  N ,
tanh

k
T
k
T
k
T
k
T
B 
B 
B
B 

 
S / NkB 1.0
0.8
S/Nk
0.692 B
ln2  0.693
B/T  0, S = NkB ln2
B/T  , S = 0
high-T
(low-B) limit
1
ln2  0.693
0.6
fi
0.4
0.2
0.0
0
2
4
x = B / kBT
6
low-T
0
(high-B)
limit
0.5
0
0
0.02
2
4
kBT/1B = x-1
x
5
x
Low-T limit
B
k BT
 1
 

B 
B  B
B 




S N,
  N k B ln 2 cosh k T   k T tanh k T 
k
T
B 
B 
B
B 

 
  e x  ex 

e x  ex 
1  e 2 x 
2 x
  x x  x   N k B  x  ln 1  e
 N k B ln 2
x
2 x 
2
e

e
1

e



 






 N k B x  e  2 x  x 1  2e  2 x  N k B e  2 x 2 x  1
S / NkB 1.0
Which x can be considered large (small)?
0.8
ln2  0.693
e. g., x  2, e 2 x  0.018  1.8%
0.6
 e 2 x  2 x  1  0.1
0.4
0.2
0.0
0
2
4
x = B / kBT
6
CB / NkB
The Heat Capacity of a Paramagnet

 B / k BT 
 U 
CB  
  N kB
cosh2  B / k BT 
 T  N , B
2
0.5
0.4
0.3
e x  e x
cosh x 
2
0.2
0.1
0.0
0
The low -T behavior: the heat capacity is small
because when kBT << 2 B, the thermal fluctuations
which flip spins are rare and it is hard for the system to
absorb thermal energy (the degrees of freedom are
frozen out). This behavior is universal for systems
with energy quantization.
The high -T behavior: N ~ N  and again, it is hard
for the system to absorb thermal energy. This behavior
is not universal, it occurs if the system’s energy
spectrum occupies a finite interval of energies.
1
2
3
kBT
kBT
NkB/2
E per particle
5
2 B
C
compare with
Einstein solid
4
kBT / B
equipartition theorem
(works for quadratic degrees
of freedom only!)
2 B
The Magnetization, Curie’s Law
N
The magnetization:
M   N   N    
B
U

 N  tanh
B
k
T
 B 
- N
 B/kBT
x  1  tanhx   x
The high-T behavior for all
paramagnets (Curie’s Law)
N  2B
M
k BT
M
N
1
 B/kBT
Negative T in a nuclear spin system  NMR  MRI
Fist observation – E. Purcell and R. Pound (1951)
Pacific Northwest National Laboratory
An animated gif of MRI images of a human head.
- Dwayne Reed
By doing some tricks, sometimes it is possible to create a metastable nonequilibrium state with the population of the top (excited) level greater than that for
the bottom (ground) level - population inversion. Note that one cannot produce a
population inversion by just increasing the system ’ s temperature. The state of
population inversion in a two-level system can be characterized with negative
temperatures - as more energy is added to the system,  and S actually decrease.
Metastable Systems without Temperature (Lasers)
For a system with more than two energy levels, for an arbitrary population of
the levels we cannot introduce T at all - that's because you can't curve-fit an
exponential to three arbitrary values of #, e.g. if occ. # = f (E) is not monotonic
(population inversion). The latter case – an optically active medium in lasers.
 E
exp  
 k BT
E4
E3



Population inversion
between E2 and E1
E2
E1
Sometimes, different temperatures can be introduced for different parts of the
spectrum.
Problem
A two-state paramagnet consists of 1x1022 spin-1/2 electrons. The component of the
electron’s magnetic moment along B is  B =  9.3x10-24 J/T. The paramagnet is
placed in an external magnetic field B = 1T which points up.
(a)
(b)
(c)
Using Boltzmann distribution, calculate the temperature at which N= N/e.
Calculate the entropy of the paramagnet at this temperature.
What is the maximum entropy possible for the paramagnet? Explain your reasoning.
spin 1/2
(two levels)
(a)
B
B
E2  E1 2 B B

1
k BT
k BT
- B
E2 = + BB
N
N
E2  E1  2 B B
B 
e
 9.3 1024 J/T
2 me
2 B B 2  9.3 1024 J/T 1 T
T

 1.35 K
 23
kB
1.3810 J/K
kBT
E1= - BB
N
 E  E1 

 exp  2
N
k BT 

Problem (cont.)
 
 B B 
B B  B B
B B 




S N,
  N k B ln 2 cosh k T   k T tanh k T 
k
T
B
B
B
B



 

 11022 1.381023 J/K  0.09  0.012J/K
If your calculator cannot handle cosh’s and sinh’s:
e x  e x
cosh x 
2
S/NkB
fi
e x  e x
sinh x 
2
1
0.5
0.09
0
0
2
4
kBT/1B
xi
e x  e x
tanhx  x  x
e e
Problem (cont.)
(b)
the maximum entropy corresponds to the limit of T   (N=N): S/NkB  ln2
B B
9.3 1024 J/T  2 T
3


4
.
5

10
k BT 1.381023 J/K  300 K
For example, at T=300K:
E2
  B
S  N , B   11022 1.381023 J/K  0.693  0.1 J/K
 k BT 
kBT
E1
ln2
S/Nk
0.692
B
1
T
S/NkB  ln2
fi
0.5
0
0
0
T  0 0.02
S/NkB  0
2
4
kBT/1 B
xi
5