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The Boltzmann Distribution
• This is the Boltzmann distribution and gives “the probability that a system in contact with a heat bath at temperature T should be in a particular state”.
• r labels all the states of the system. At low temperature only the lowest states have any chance of being occupied. As the temperature is raised higher lying states become more and more likely to be occupied.
• In this case, in contact with the heat bath, all the microstates are therefore not equally likely to be populated.
kT
U
r
r
eZ
p
1
r
kT
Ur
eZ
The Boltzmann Distribution
• Usually there are huge numbers of microstates that can all have the same energy. This is called degeneracy.
• In this case we can do our summations above over each individual energy level rather than sum over each individual microstate.
• The summation is now over all the different energies Ur and g(Ur) is the number of states possessing the energy Ur. The probability is that of finding the system with energy Ur.
kT
U
r
r
eZ
p
1
r
kT
Ur
eZ
kT
U
rr
r
eUgZ
Up
)(1
)(
r
r
U
kT
U
r eUgZ )(
Entropy in ensembles
• Our system embedded in a heat bath is called a canonical ensemble (our isolated system on its own from Lecture 16 is termed a microcanonical ensemble).
• When isolated the microcanonical ensemble has a defined internal energy so that the probability of finding a system in a particular microstate is the same as any other microstate.
• In a heat bath the energy of the system fluctuates and the probability of finding any particular microstate is not equal. Can we now calculate the entropy for such a system and hence derive thermodynamic variables from statistical properties?
Entropy in the canonical ensemble• Embed our system in a
heat bath made up of (M-1) replica subsystems to the one were interested in.
• Each subsystem may be in one of many microstates. The number of subsystems in the ith microstate is ni.
• The number of ways of arranging n1 systems of µstate 1, n2 systems of µstate 2, n3….
!
!
iin
MW
Entropy in the canonical ensemble
• This is the general definition of entropy and holds even if the probabilities of each individual microstate are different.
• If all microstates are equally probable pi= 1/W (microcanonical ensemble)
• Which brings us nicely back to the Boltzmann relation
i
ii ppkS ln
WkWW
kppkSW
iiii ln
1ln
1ln
1
Entropy in the canonical ensemble
• The general definition of entropy, in combination with the Boltzmann distribution allows us to calculate real properties of the system.
i
ii ppkS ln
i
ii
iii Z
kT
UpkppkS lnln
kT
U
i
i
eZ
p
1
i
kT
Ui
eZ
ZkT
Up ii lnln
ZkT
UpZkUp
TS
ii
iii lnln
1
Helmholtz Free Energy
• Ū is the average value of the internal energy of the system.
• (Ū – TS) is the average value of the Helmholtz free energy, F. This is a function of state that we briefly mentioned in earlier lectures. It is central to statistical mechanics.
• The Partition function Z has appeared in our result –it seems to be much more than a mere normalising factor. Z acts as a bridge linking the microscopic world of microstates (quantum states) to the free energy and hence to all the large scale properties of a system.
ZkT
US ln ZkTTSUF ln
Helmholtz Free Energy
• F is a state function. Can we now calculate the thermal properties of our system?
• Lets ignore Ū is an average and assume U = Ū.
• We can now use our thermodynamics knowledge to define our thermodynamic variables. For a small reversible change:
ZkTTSUF ln
TSUF
SdTTdSdUdF
SdTTdSPdVdQdF R
SdTPdVSdTTdSPdVTdSdF
Helmholtz Free EnergyZkTTSUF lnSdTPdVdF
• Employing partial differentials we find:-
• We have intimately related the energies of the microstates of the system to the pressure and entropy.
PV
F
T
TT V
ZTk
V
FP
ln
ST
F
V
VV T
ZTk
T
FS
ln
Response functions –2nd derivatives of free energy
• Moduli of elasticity are the stress/strain or force/unit area divided by the fractional deformation (Lecture 8):-
• Most usefully the heat capacity at constant volume:-
TT V
FV
V
PVK
2
2
VVV
V T
FT
T
ST
T
QC
2
2
Mean Energy• Ū is the average value of the internal
energy of the system. The actual internal energy fluctuates because we have defined the temperature of the heat bath.
• How big are the fluctuations? – are they important?
VT
ZTkTZkTTSFU
)ln(
ln
T
ZkT
T
ZTZkTZkTU
V
lnln
lnln 2
Fluctuations in Internal Energy• We measure departures from the mean
value by use of standard deviations - as we would with any distribution.
2222)( UUUUU
Vr
kT
Ui
kT
U
i
V
V r
i
e
eU
TT
UC
Fluctuations in Internal Energy
2222)( UUUUU
22
2
1UU
kTe
eU
TC
Vr
kT
Ui
kT
U
i
V r
i
VCkTUUU 2222)(
Fluctuations in Internal Energy• We have calculated the variance – the relative
fluctuation U/ Ū is of most use:-
• But Ū and CV are extensive properties proportional to the size of the system ~ N, the number of particles in the system. T intensive and is size independent.
U
CkT
U
U V
2
NU
U 1
Fluctuations in Internal Energy• For typical macroscopic systems with ~1023 particles fluctuations U/
Ū ~ 10-11
• Fluctuations are tiny and hence U and Ū can be considered identical for all practical purposes.
• Macroscopic systems in a heat bath effectively have their energy determined.
• Similar relationships can be found for other relative fluctuations of properties of macroscopic systems.
NU
U 1
Summary – Statistical Mechanics• Microstate – The state of a system defined
microscopically – a complete description on the atomic scale.
• Macrostate – The state of a stystem of macroscopic size specified by a few macroscopically observable quantities only.
• Statistical Weight (W or ) of a macrostate – is the number of microstates compising the macrostate.
• Postulate of equal a priori probabilities – “for an isolated system in a definite macrostate, the W microstates comprising this macrostate occur with equal probability.
• Equilibrium Postulate – “for an isolated macroscopic system, defined by U,V,N (which are fixed) and variable parameters , equilibrium corresponds to those values of for which the statistical weight W(U,V,N, ) attains its maximum.”
Summary – Statistical Mechanics• Boltzmann definition of the Entropy
• Temperature definition
• Pressure definition
• General definition of the Entropy
),,,(ln),,,( NVUWkNVUS
NUV
NVUSTP
,
),,(
i
ii ppkS ln
NVU
NVUS
T ,
),,(1
Summary – Statistical Mechanics• Boltzmann Distribution – pi is the probability that a
system at temperature T is in the state i with energy Ui.
• Partition Function – summed over all microstates
• Mean Energy
• Helmholtz free energy F = U - TS
kT
U
i
i
eZ
p
1
T
ZkTU
ln2
i
kT
U i
eZ
Paramagnetic materials
• This is a simple model to develop our understanding of stat. Mechanics but it proves to be very significant.
• Paramagnets contain atoms which have magnetic dipole moments (). These do not interact with each other but can respond to an applied external magnetic (B) field.
• The dipoles can be thought of as independent (atomic) bar magnets arranged in a crystal lattice.
Paramagnetic materials
• The dipoles can be thought of as independent (atomic) bar magnets arranged in a crystal lattice.
• Think
• In a B field each dipole can exist in one of two states – aligned with the field (spin up) or anti aligned (spin down).
• Spin up dipoles have an energy -B, spin down +B.
• We want to find out how the magnetisation of the material depends on temperature and applied field.
Paramagnetic materials
• As all the dipoles are independent of each other we really only need to look at the average properties of one dipole. We can use all the other dipoles as the heat bath – Canonical ensemble.
• We have the two possible microstates and energies already – get Partition Function Z1 for our single dipole.
xkT
BeeeZ kT
B
kT
B
i
kT
U i
cosh2cosh2
kT
Bx
Paramagnetic materials• We can now calculate the probability of
spin up versus spin down.
kT
U
i
i
eZ
p
1 xe
xp
cosh2
1 xex
p
cosh2
1
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Pro
ba
bilit
y
X=B/kT
Spin aligned (U = -B) Spin antialigned (U = +B)
Paramagnetic materials• We can now calculate the mean magnetic
moment of our individual dipole.
• And hence the mean energy of the individual
xex
p cosh
1 xex
p cosh
1
xx
eex
pp xx sinh2cosh2
)(cosh2
xtanh
xBU tanh
Paramagnetic materials• We have the values necessary for the individual
dipole and because our dipoles do not interact all other dipoles must behave similarly.
• A solid of N dipoles therefore has and energy:
• And a mean magnetic moment (in the direction of the applied field) of:
• Note that U = -MB
xNNM tanh
xNBUNU tanh
Paramagnetic materials• The magnetisation L is the magnetic
moment per unit volume
• In the limit of a weak filed or high temperature x<<1 and tanh x x so:
V
xN
V
ML
tanh
BVkT
N
V
NxL
2
Paramagnetic materials
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Spin Up Spin Down Magnetisation
X=b/kT
Pro
bab
ility
0.0
0.2
0.4
0.6
0.8
1.0
Mag
netisa
tion
/ N/V
Paramagnetic materials –Curie’s Law• The susceptibility is the magnetisation per applied
field intensity which for small magnetisations is given by H = B/0.
• Is Curie’s law. It holds very well for paramagnetic materials with weakly interacting dipoles. So well that it can be used for temperature calibration for example Cerium Magnesium nitrate obeys curies lay to 0.01K!
VkT
N
H
L 0
2
T
1
Paramagnetic materials
• The dipoles can be thought of as independent (atomic) bar magnets arranged in a crystal lattice.
• Think
• In a B field each dipole can exist in one of two states – aligned with the field (spin up) or anti aligned (spin down).
• Spin up dipoles have an energy -B, spin down +B.
• We want to find out how the magnetisation of the material depends on temperature and applied field.
Paramagnetic materials
• As all the dipoles are independent of each other we really only need to look at the average properties of one dipole. We can use all the other dipoles as the heat bath – Canonical ensemble.
• We have the two possible microstates and energies already – get Partition Function Z1 for our single dipole.
xkT
BeeeZ kT
B
kT
B
i
kT
U i
cosh2cosh2
kT
Bx
Paramagnetic materials• We have the values necessary for the individual
dipole and because our dipoles do not interact all other dipoles must behave similarly.
• A solid of N dipoles therefore has an energy:
• And a mean magnetic moment (in the direction of the applied field) of:
• Note that U = -MB
xNNM tanh
xNBUNU tanh
Paramagnetic materials
0.0 0.5 1.0 1.5 2.0 2.5 3.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Spin Up Spin Down Magnetisation
X=b/kT
Pro
bab
ility
0.0
0.2
0.4
0.6
0.8
1.0
Mag
netisa
tion
/ N/V
Hyperbolic Functions – Flap M4.6!• Reminder – please revise
xx
xx
ee
ee
x
xx
)cosh(
)sinh()tanh(
2)cosh(
xx eex
2)sinh(
xx eex
)cosh(2
)sinh( xee
xdx
d xx
)sinh(2
)cosh( xee
xdx
d xx
Hyperbolic Functions – Flap M4.6!
xx
xx
ee
ee
x
xx
)cosh(
)sinh()tanh(
)(sech)(cosh
1)tanh( 2
2x
xx
dx
d
Use quotient rule to prove
Heat Capacity
• Our paramagnetic solid has an energy that depends upon temperature – it therefore must have a magnetic heat capacity.
• Experimentally one measures the heat capacity at constant magnetic field intensity H.
• as our paramagnetic compound is weakly magnetic H = B/0 so B is also constant.
kT
BNBxNBUNU
tanhtanh
kT
B
TNB
T
U
T
dQC
HH
H
tanh
Heat Capacity
kT
B
kT
BNk
kT
B
TNBCH
2
2
sechtanh
0 1 2 3 4 5
0.0
0.2
0.4
CH /
Nk
1/X = kT/B
Magnetic Heat Capacity
Schottky Heat Capacity
kT
B
kT
BNkCH
2
2
sech
0 1 2 3 4 5
0.0
0.2
0.4
CH /
Nk
1/X = kT/B
Magnetic Heat Capacity
• This is in fact a general result for the heat capacity in any two level system. One example we have already encountered is the Schottky defect.
Isolated Paramagnetic Solid• Very similar problem to the one previously treated in a
heat bath. Here we constrain (fix) the total energy U of the isolated system.
• N total dipoles, n spin-up aligned with the applied B field • U is clearly a function of n.
• A given energy U(n) corresponds to a given number of n spin up atoms with statistical weight:-
)2()()( nNBBnBnNnU
!)!(
!)(
nnN
NnW
Isolated Paramagnetic Solid• Hence Entropy is given by:-
• For large N (~1023) can use Stirling’s Approximation
• Should look familiar – it’s the same problem as the Schottky Vacancy formation.
!)!(
!ln)(ln)(
nnN
NknWknS
)ln()(lnln)( nNnNnnNNknS
n
nS
dU
dn
n
nS
nU
nS
U
S
T
)(1)(
)(
)(1
Isolated Paramagnetic Solid
• Solve for n, to find the density of spin up atoms:-
• This is identical to the probability of finding a spin up atom in a heat bath we started with last lecture!
n
nN
B
k
n
nS
BTln
2
)(
2
11
xeZN
n
1
1
dU
dn
n
nS
nU
nS
U
S
T
)(
)(
)(1)2()( nNBnU
Negative Temperature• From our previous derivation we had
• If n < N/2 then more than half the dipoles are anti-parallel 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.
nN
n
B
k
n
nN
B
k
Tln
2ln
2
1
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=NB.• 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=-NB and the highest U=+NB. These are both unique microstates so S=0.
• In between we can only reach states with positive energy with a negative temperature.
-5 -4 -3 -2 -1 0 1 2 3 4 5
-1.0
-0.5
0.0
0.5
1.0
En
erg
y /
NB
Temperature
System Energy
-20 -10 0 10 20
-1.0
-0.5
0.0
0.5
1.0
En
erg
y /
NB
1 / Temperature
System Energy
