Transcript Lecture 7. Systems with a “Limited” Energy Spectrum

Lecture 8. 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