Download Thermal Physics PH2001

Dr Roger Bennett
[email protected]
Rm. 23 Xtn. 8559
Lecture 22
Negative Temperature
• From our previous derivation we had
1  k  N  n
k
 n 

ln 
ln 


T 2 B  n  2 B  N  n 
• If n < N/2 then more than half the dipoles are antiparallel and T becomes negative!
• What is a negative temperature?
• We know that as the temperature T the populations
of spin-up and spin-down only become equal!
• A negative temperature state must therefore be hotter
than T= as its is a more energetic state of the system.
Negative Temperature
• For a negative temperature the entropy and statistical
weight must be decreasing functions of E.
• This can happen if the system possess a state of finite
maximum energy – such as our paramagnet with
U=NB.
• No systems exist where this happens for all particular
aspects (I.e. vibrational energies, electronic energies and
magnetic energies). However, if one such aspect or
subsystem is effectively decoupled from the others, so
they do not interact, that subsystem may be considered
to reach internal equilibrium without being in equilibrium
with the others.
• This is the case for magnetic systems where the
relaxation times between atomic spins is much quicker
than the relaxation between spins and the vibrational
modes of the lattice.
Negative Temperature
• In the paramagnet the lowest possible energy is
U=-NB and the highest U=+NB. These are
both unique microstates so S=0.
• In between we can only reach states with
positive energy with a negative temperature.
1.0
1.0
0.5
Energy / NB
Energy / NB
System Energy
System Energy
0.5
0.0
-0.5
-1.0
0.0
-0.5
-1.0
-5
-4
-3
-2
-1
0
1
Temperature
2
3
4
5
-20
-10
0
1 / Temperature
10
20
Paramagnets – Adiabatic cooling
• We can use a paramagnet to cool a sample by cycling
the magnetic fields and allowing or blocking heat
exchange.
• If the B field is reduced while keeping magnetic
entropy constant the temperature must fall to keep
the same degree of order.
• To do this we need to find the entropy and how that
depends on temperature in this system.
• We know U, T is our variable and we want to find S.
• What connects all these?
Paramagnets – Adiabatic cooling
• Helmholtz free energy F = U-TS
U  N U   NB tanh x   NB tanh
B
kT
F  U  TS  kT ln Z  kNT ln Z1
Z1   e
i
U i
kT
e
 B
kT
e
 B
kT
 B 
 2 cosh
  2 cosh x
 kT 
U F
S
 Nk ln 2  ln(cosh x )  x tanh( x )
T
Paramagnets – Adiabatic cooling
• Paramagnet in contact with a heat bath
• Apply and external B field – entropy is
reduced as spins align. Temperature fixed by
heat bath.
• Heat bath removed –sample now isolated
• Magnetic field slowly reduced reversibly – not
quite to zero because of residual alignment of
spins (Bapplied~100 Bresidual).
• No heat flows into the system so ordering
remains same.
Paramagnets – Adiabatic cooling
U F
S
 Nk ln 2  ln(cosh x )  x tanh( x )
T

A
0 .7
Low Field
High Field
0 .6
A

S / Nk
0 .5
0 .4
C
0 .3
B
0 .2
B
2B
0 .1
A-B Is oth ermal ma gn etisatio n
B-C Re vers ib le ad iab atic de mag ne tisa tion (slo w)
0 .0
- 0 .1
0
1
Temperature
2
3

C
Paramagnets – Adiabatic cooling
S  Nk ln 2  ln(cosh x )  x tanh( x )
• How big an effect is it?
• S is a function of B / T only
• S remains constant during
adiabatic reversible
demagnetisation.
• Ratio B/ T therefore constant.
Bresidual Tinital
T final  Tinital *

Bapplied 100
• Can use nuclear spins to
T=10-6
x
B
kT

A

B
K

C
Dr Roger Bennett
[email protected]
Rm. 23 Xtn. 8559
Lecture 23
Quantum statistics
•
1.
2.
3.
4.
5.
In lectures and workshop we have considered model
systems to develop/predict thermodynamic properties.
Don’t focus on the mathematics – think of it as a
systematic method to getting to the thermodynamic
properties of the system.
Create a simple model of the system
Identify the energetic states of the system
Calculate the partition function from the energy levels.
Deduce the free energy
Differentiate to get the entropy, pressure use a 2nd
derivative to get response functions –Heat capacity and
elastic moduli.
6. Compare with real data – is it a good fit – refine model.
Quantum statistics
•
So far we have taken a classical viewpoint where
particles are distinguishable – think about the way we
labelled lattice sites in a crystal or position along a
chain of paramagnetic atoms.
•
Quantum mechanics complicates matters slightly as
particles cannot be localised and the wavelike nature
results in problems in identifying unique quantum
microstates.
•
We will now start on this problem – safe in the
knowledge that we already have the tools to get to the
thermodynamics if we can just get the quantum states
and partition function right.
Quantum statistics
•
•
•
We shall start with a quantum particle in a box.
We replicate the box (M-1) times to make the
entire system. We work as previously for the
Canonical Ensemble.
We localise each particle in the box and then
make each box distinguishable (i.e. we can
label the boxes).
As before to determine the thermodynamics we
need to list the independent quantum states,
determine the energies and calculate the
partition function.
Quantum statistics
•
•
Simple problem first imagine a 1-D potential well
of width L with zero
potential inside and infinite
potential defining the walls
located at x=0 and x=L.
Schrödingers wave equation
gives us the quantum states
that are allowed


 2  2 ( x )

 V ( x ) ( x )   ( x )
2
2m x
L
0
Quantum statistics
•
•
Simple problem first - imagine a 1-D potential well of
width L with zero potential inside and infinite potential
defining the walls.
Schrödingers wave equation gives us the quantum

states that are allowed

 2  2 ( x )

 V ( x ) ( x )   ( x )
2
2m x
 nx 
n ( x )  A sin 

 L 
•
0
Boundary conditions satisfied provided n is a positive
non zero integer.
Quantum statistics
•
Schrödingers wave equation gives us the
quantum states that are allowed. From the
wavefunction we also get the energy

eigenvalues.
   ( x)

 V ( x ) ( x )   ( x )
2
2m x
 nx 
n ( x )  A sin 

 L 
2
2
  n 
2
n  


n

2
2
mL


2
2
2
0

Quantum statistics
 nx 
n ( x )  A sin 

 L 
•
•
  2 2 n2 
2
n  


n

2
 2mL 
The energy scale is set by . For an electron in a 1cm
box  ~ 6×10-34 J – pretty small in comparison to
room temperature where kT = 4×10-21 J (~25meV).
We now have the number of allowed states and their
energies –we can proceed to find the partition
function for our single particle in the box.
Quantum statistics
  n 
2
Z1 
n  


n

2
 2mL 
2 2

 

 

2
 2kTmL 
2
•
•
2
2

e
 n
kT
n1

 e
n2
n1
We have an infinite series with the factor 
very small so it converges very slowly.
Because  is very small we can approximate
the sum to an integral:

Z1   e
n1
n

2
 e
0
n2
dn
Quantum statistics
•
This is almost the standard integral:
 
0 e dx   4 

•
 x2
Change variable
1
2
x n
1
 
 L mkT 
n
Z1   e dn     

2
0
 2π 
 4 

2
2
1
2
2
•
Can now use our existing framework to calculate
thermodynamics
 L mkT 
Z1  

2
 2π 
Quantum statistics
•
2
Find the free energy of the single particle
1
2
2
L
mkT

kT
L


 mkT 
F   kT ln Z   kT ln 
ln 
 

2
2
2
 2π 
 2π 
2
•
•
Differentiate w.r.t. T to find entropy and again for Cv
2


F
k
L
mkT  



S  
  1
   ln 
2
 T V 2   2π  
k
 S 
CV  T 
 
 T V 2
The translational motion of a single particle in 1-D
gives this contribution to Cv
1
2
1-D Quantum summary
1. Create a simple model of the system
– Identical 1-D wells each with one particle
2. Identify the energetic states of the system
– Eigenstates of the well with defined energy
eigenvalues
3. Calculate the partition function from the energy levels
– Summation becomes an integral “do the math!”
4. Deduce the free energy
– Remember F = -kTln(Z)
5. Differentiate to get the entropy, pressure use a 2nd
derivative to get response functions – Heat capacity and
elastic moduli.
6. Compare with real data – is it a good fit – refine model.
– Looks OK – what about 3-D real systems?
Dr Roger Bennett
[email protected]
Rm. 23 Xtn. 8559
Lecture 24
1-D Quantum summary
1. Create a simple model of the system
– Identical 1-D wells each with one particle
2. Identify the energetic states of the system
– Eigenstates of the well with defined energy
eigenvalues
3. Calculate the partition function from the energy levels
– Summation becomes an integral “do the math!”
4. Deduce the free energy
– Remember F = -kTln(Z)
5. Differentiate to get the entropy, pressure use a 2nd
derivative to get response functions – Heat capacity and
elastic moduli.
6. Compare with real data – is it a good fit – refine model.
– Looks OK – what about 3-D real systems?
1-D Quantum statistics – single particle
•
The free energy of the single particle.
 kT  L2 mkT 
F   kT ln Z 
ln 

2
2
 2π 
•
The entropy.
2

k  L mkT  
 F 
S  
  1
   ln 
2
 T V 2   2π  
•
The constant volume heat capacity Cv.
k
 S 
CV  T 
 
 T V 2
3-D Quantum statistics – single particle
•
•
•
We can use the same methodology to extend our
analysis to 3D.
Imagine our single particle now trapped in a cube of
side L aligned with the X,Y,Z axis for convenience.
The three directions are independent so we can
simply write the wave function as the product of the
separate functions.
 n1x   n2y   n3z 
i ( x, y, z )  A sin 
 sin 
 sin 

 L   L   L 
•
This is a standing wave in three dimensions where
the subscript i on the wave function signifies a unique
set of quantum numbers (n1,n2,n3) – a microstate.
3-D Quantum statistics – single particle
•
The energy of the free particle of mass m can be
found from the Schrödingers wave equation


2i ( x, y, z )  V ( x, y, z )i ( x, y, z )   ii ( x, y, z )
2m
2
•
The boundary conditions of infinite potential bounding
the box with zero potential inside dictate that n1, n2
and n3 are all positive, non zero, integers.
  2 2  2
2
2
i  
( n1  n2  n3 )
2
 2mL 
3-D Quantum statistics – single particle
•
From these energies we can construct the
partition function for the translational motion
2 2

   2
2
2
i  
(
n

n

n
)

1
2
3
2
 2mL 

ZTrans   e
i 1
 i
kT



   e
n1 1 n2 1 n3 1
  2 2 
 

2
 2kTmL 
 ( n12  n22  n32 )
3-D Quantum statistics – single particle

ZTrans   e
 i
kT
i 1

  e
 n 1

ZTrans


   e
 ( n12  n22  n32 )
n1 1 n2 1 n3 1
n12
1
•


  e
 n 1

n22
2

n 
  e 
 n 1


2
3
3
We can solve these summations as per the 1-D case –
same assumptions apply.
3
ZTrans
 mkT 
 mkT 
Z L
V
2 
2 
 2 
 2 
3
1
3
2
3
2
3-D Quantum statistics – single particle
ZTrans
•
 mkT 
V
2 
 2 
3
2
Once we have the partition function – get free energy
3  mkT  

F   kT ln Z  kT ln(V )  ln 
2 
2  2π  

•
And the pressure
kT
 F 
P  
 
 V T V
3-D Quantum statistics – single particle
kT
 F 
P  
 
 V T V
•
•
PV  kT
This should be familiar! It’s the ideal gas law for a
single particle. If we have N particles it becomes
PV=NkT provided the particles do not interact.
And the entropy and heat capacity
3  mkT  3 

 F 
S  
 
  k  ln(V )  ln 
2 
2  2π  2 
 T V

3k
 S 
CV  T 
 
 T V 2
3-D Quantum statistics – summary
•
We recover the ideal gas law from first principles
under the assumption the particles do not interact.
•
We have to take care about multiplying by N to scale
up the single particle result to many particles. It
works for the heat capacity but not for free energy or
the entropy. To do it properly we need to find the
partition function for an N atom quantum system.
kT
 F 
P  
 
 V T V
PV  NkT
3  mkT  3 

S  k  ln(V )  ln 
 
2 
2  2π  2 

3k
 S 
CV  T 
 
 T V 2
Expressions for heat and work
• We have a model that seems to give us the ideal gas
behaviour – can we link processes variables like heat
and work to the microscopic details.
• The average internal energy is
U   pi Ei
i
• If in a heat bath then the probabilities are given for
each quantum (micro-) state I with energy eigenvalue
 Ei
Ei.
kT
Pi 
• For an infinitesimal quasistatic process
dU   dpi Ei   pi dEi
i
i
e
Z
Expressions for heat and work
• We already have a thermodynamic meaning of these
terms.
dU = đQR + đWR
dU   dpi Ei   pi dEi
i
i
• Which term is which?
• If we fix all external parameters such as volume,
magnetic fields etc we fix the positions of the energy
levels as they only depend on those parameters. So
đEi=0. In this case the only way to increase energy is
to add heat to the system.
dQ   dpi Ei
i
Expressions for heat and work
• It follows that the work done on the system is given by.
dW   pi dEi
i
• Doing work is the same as the weighted average
change of the energy levels.
dQ   dpi Ei
E3
i
E3
E2
P3
dW   pi dEi
P3
i
P2
E1
E2
E1
P2
P1
Initial State
P1
E3
E2
E1
P3
P2
P1
Dr Roger Bennett
[email protected]
Rm. 23 Xtn. 8559
Lecture 24
3-D Quantum statistics – summary
•
We recover the ideal gas law from first principles
under the assumption the particles do not interact.
•
We have to take care about multiplying by N to scale
up the single particle result to many particles. It
works for the heat capacity but not for free energy or
the entropy. To do it properly we need to find the
partition function for an N atom quantum system.
kT
 F 
P  
 
 V T V
PV  NkT
3  mkT  3 

S  k  ln(V )  ln 
 
2 
2  2π  2 

3k
 S 
CV  T 
 
 T V 2
Multiple particles- what’s the problem?
•
Lets re-visit the partition function for the single
particle in a 3-D box.
3
ZTrans
•
 mkT 
V
 VnQ
2 
 2 
2
Where we have defined a new quantity nQ which is
the quantum concentration. Since the particle
concentration for this single atom case is 1/V we see
that Ztrans is the ratio of the quantum concentration to
the particle concentration.
Quantum concentration
•
The quantum concentration is the concentration
associated with a particle in a cube whose length of
side is given (roughly) by the thermal average de
Broglie wavelength.
3
 mkT 
3
nQ  



2
 2 
•
2
For helium at room temperature the atomic
concentration n = 1/V ~2.5  1019 cm-3 and quantum
concetration nQ ~0.8  1025 cm-3. Thus n/nQ ~10-6
which makes He very dilute under normal conditions.
Quantum concentration
•
If the system under consideration is very dilute I.e.
nQV = nQ/n<<1 then the quantum mechanical “size”
of the particle is much smaller than the box its
effectively trapped in.
•
If nQV = nQ/n<<1 the gas may be considered to be in
the classical regime and quantum effects can be
neglected.
3
 mkT 
3
nQ  



2
 2 
2
Mean energy of the particle
• We previously established (L19) that the mean
energy of a subsystem in contact with a heat bath
was given by:
 ln Z
U  kT
T
2
• We have Z for our single atom
ZTrans
 mkT 
V
2 
 2 
3
2
2
2

ln
Z
kT

Z
kT
3Z trans 3kT
2
trans
trans
U  kT



T
Z trans T
Z trans 2T
2
N atoms (at last!)
• Lets extend our model to N atoms in the box. We’ll
do this in 2 stages: first we assume we can
distinguish between the atoms somehow.
• N distinguishable particles in the box that do not
interact with each other the system energy is the
sum of their individual energies.
• The partition function is the sum of the Boltzmann
factors over every possible state of the system (as
always - this isn't new).
• These states can be found by taking every possible
state of atom 1 with every possible state of atom 2
with every possible state of all the other atoms……
N distinguishable atoms
• These states can be found by taking every possible state of
atom 1 with every possible state of atom 2 with every possible
state of all the other atoms……

Z dist   e
i 1

Z dist   e
i 1
 i
kT

 e
 (  atom1  atom2  atom3 ....) i
kT
i 1
(  atom1  atom2  atom3 ....) i
kT

 e
 i 1

• The partition function of the system is the product of the
partition functions for the individual particles.
 i
kT



N
N
Z dist   Z i
i 1
N distinguishable atoms - summary
• The partition function of the system is the product of
the partition functions for the individual particles.
N
Z dist   Z i
i 1
• This implies the free energy is is extensive:N
N
i 1
i 1
F  kT ln Z dist  kT  ln Z i   Fi
• If our particles were all distinguishable but had the
same single particle partition function then the
partition function for the system would be:
Z dist  Z i
N
N indistinguishable atoms
• 2nd stage N indistinguishable particles in the box that do
not interact with each other the system energy is the
sum of their individual energies.
• The partition function is the sum of the Boltzmann factors
over every possible state of the system (as always - this
isn't new).
• However, as we now have indistinguishable particles we
massively over count the number of distinct states.
• If it can be assumed that the number of available states
is much larger than the number of particles then the
probability of finding any two particles in the same state
is very low. We then have N! different ways of assigning
those states to the particles. But as the particles are
indistinguishable all of these would be the same state!
N indistinguishable atoms
• We then have N! different ways of assigning those states
to the particles. But as the particles are indistinguishable
all of these would be the same state!
• We have over-counted by a factor of N! for which we can
correct:N
Zi
ZN 
N!
• The assumption that the number of states is far greater
than the number of particles is true if we are in the
classical regime nQ>>n.
• This adjustment is called corrected classical counting.
N indistinguishable atoms
• Energy of an N particle gas.
N


Z trans 
2  ln Z N
2 

U  kT
 kT
ln 
T
T   N ! 
3
2 
 kT
( N ln Z trans  ln N ! )  NkT
T
2
• Free energy of N particle gas (using Stirling’s approx.)
F  kT ln Z N   NkT ln Z trans  kT ( N ln N  N )
N indistinguishable atoms
• Pressure of an N particle gas.
1 Z trans NkT
 F 
P  

  NkT
Z trans V
V
 V T ,N
• Entropy of N particle gas the Sackur-Tetrode equation
kT Z trans
 F 
S  
 k ( N ln N  1)
  kN ln Z trans 
Z trans T
 T V ,N
  n
3
  Q


 Nk ln Z trans   ln N  1  Nk ln
N
2


  V
 5 

 2


Dr Roger Bennett
[email protected]
Rm. 23 Xtn. 8559
Lecture 25
Quantum statistics
1. Create a simple model of the system
2. Identify the energetic states of the system
3. Calculate the partition function from the energy
levels.
4. Deduce the free energy
5. Differentiate to get the entropy, pressure use a
2nd derivative to get response functions –Heat
capacity and elastic moduli.
6. Compare with real data – is it a good fit –
refine model.
Quantum concentration
•
If the system under consideration is very dilute i.e.
nQV = nQ/n<<1 then the quantum mechanical “size”
of the particle is much smaller than the box its
effectively trapped in.
•
If nQV = nQ/n<<1 the gas may be considered to be in
the classical regime and quantum effects can be
neglected.
3
 mkT 
3
nQ  



2
 2 
2
N distinguishable atoms
• These states can be found by taking every possible
state of atom 1 with every possible state of atom 2
with every possible state of all the other atoms……

Z dist   e
(  atom1  atom2  atom3 ....) i
kT
i 1

 e
 i 1

 i
kT



N
• The partition function of the system is the product of
the partition functions for the individual particles.
• If our particles have internal structure (molecules)
then we can separate centre of mass motion (solved
already) from internal motions. Etotal = Etrans + Einternal
N
Z dist  ( Z int Z trans )
i 1
N
Z dist   Z i
i 1
N indistinguishable atoms
• We then have N! different ways of assigning those states
to the particles. But as the particles are indistinguishable
all of these would be the same state!
• We have over-counted by a factor of N! for which we can
correct:N
Zi
ZN 
N!
• The assumption that the number of states is far greater
than the number of particles is true if we are in the
classical regime nQ>>n.
• This adjustment is called corrected classical counting.
N indistinguishable atoms
• Pressure of an N particle gas.
1 Z trans NkT
 F 
P  

  NkT
Z trans V
V
 V T ,N
• Entropy of N particle gas the Sackur-Tetrode equation
  n  5

F
  Q 



S  
  Nk ln  N   
 T V ,N
  V  2 
What’s the difference?
• Compare the Sackur-Tetrode equation
  n  5
  V  3  mkT  5 
  Q 

S  Nk ln
   Nk ln    ln 
 
2 
N  2
  N  2  2  2 
  V 

• To the entropy of a single gas particle in a volume V
3  mkT  3 

S  k ln V   ln 
 
2 
2  2  2 

• The corrected classical counting ensures the ratio V/N
appears in the entropy. This term is constant as the size
of the system is scaled and ensures entropy is extensive.
Example – Ne
• Ne (m=20.18 amu) at its boiling point (27.2K) under
1atm (1.013 105 Pa)
  n  5
  V  3  mkT  5 
  Q 

S  Nk ln
   Nk ln    ln 
 
2 
N  2
  N  2  2  2 
  V 

• For 1 mole PV=NakT, Na/V = P/kT = 2.70 1026 m-3
• nQ = 2.417 1030 m-3
• ln (nQ/(N/V)) = ln(8951) = 9.1 (c.f. 5/2)
• S=Nak[ln (nQ/(N/V)) +5/2] = 96.45 J mol-1 K-1
• Measured value S = 96.40 J mol-1 K-1
Non- examples! Not dilute
•
•
•
•
•
Liquid He
From the density of the liquid we find n=N/V= 2 1028 m-3
At 10K the de Broglie wavelength ~ 4 10-10 m
So nQ = 1.56 1028 m-3
nQ/n<<1 is therefore not true – it’s a truly quantum
system as the atoms overlap
• Conduction electrons in a metal – assume one electron
per atom so N/V ~ 1.25 10-10 1029 m-3
• This is equivalent to a box of side 210-10m
• With electrons of mass only 9.110-23kg, 210-10m as a
thermal average de Broglie length corresponds to 3105K
The Maxwell velocity distribution
• We now have a working appreciation of ideal gases from
first principles - revisit the velocity distribution. We want
to find n(u)du – the number of atoms with speeds
between u and du.
• Consider a gas of N particles enclosed in a volume V in
thermal equilibrium at temperature T.
• Probability of particular atom is in a micro (q.m) state with
speed u (and hence kinetic energy ½mu2) is:-
pu 
e
u

kT
Z1

e
 mu
2
2 kT
Z1
3
 mkT  2
Z1  V 
 VnQ
2 
 2 
The Maxwell velocity distribution
• Probability of any atom is in this micro (q.m) state is Npu
• What we need to find is the number of states for each
atom which have speeds in the desired range u to u+du.
• Recall
 n1x   n2y   n3z 
i ( x, y, z )  A sin 
 sin 
 sin 

 L   L   L 
• Q.M. doesn’t deal with velocities but momenta. A
momentum measurement in the x direction would yield
±ħkx where:
kx 
n1
Lx
ky 
n2
Ly
kz 
n3
Lz
The Maxwell velocity distribution
• As these are directional we can construct a wave-vector
k  iˆkx  ˆjk y  kˆkz
kx 
n1
ky 
Lx
n2
Ly
kz 
n3
Lz
• This is reciprocal or momentum space and will crop up
over and over again. For each solution of the wave
equation defined by integer values of (n1, n2, n3) there is a
unique state and hence point in k-space spaced apart by a
length /L – each point occupies volume = (/L)3 = 3/V .
• The number of states with magnitudes between k and
k+dk would therefore be (the 1/8 comes from only +ve n1,
n2, n3):
18 4k dkV  
2
3
The Maxwell velocity distribution
• This defines a density of states in momentum space:
2
Vk
f ( k )dk  2 dk
2
• We want speeds in real space so use de Broglie relation
| momentum| mu 
k
u
m
h

 k

du  dk
m
3
V m 2
g ( u)du  2   u du
2   
The Maxwell velocity distribution
• The number of particles in this range is therefore
• The number of states in that range × probability of an atom in such a
state.
 mu
2
3
V m 2
n( u)du  N
u du

3
2 
 mkT  2 2   
V
2 
 2 
e
2 kT
3
2  m  2 2 mu
n( u)du  N

 ue
  kT 
2
2 kT
du
• Note no ħ in the expression – it’s a classical result that we derived in
lecture 4!