Physics 112: Thermodynamicsrogerg/education/112_formulas.pdf · 2008-05-06 · about systems in...
179
Physics 112: Thermodynamics Served by Roger Griffith Nutritional Facts: Serving size: 1 Semester (16 weeks) Servings per container: many problems and solutions office hours for John Clarke Wed 3-4 349 Birge Preamble for this course Why do we need statistical mechanics? • Quantum Mechanics: Particle in a box, Harmonic oscillator, H-atom. The SE gives us our solutions. |ψ| 2 gives us the probability of finding a particle somewhere with some speed. But this is not statistics. But for a cm 3 of air ~ 3 × 10 19 molecules, this is a hopeless problem to solve the SE. But for a macroscopic system we must use a statistical approach. e.g to find the heat capacity or paramagnetic susceptibility. Statistically this turns out to yield exceedingly accurate results. Road Map 1. Define states of the system, probability, binomial distribution, gaussian, and mean value theorem. 2. Entropy: The measure of the number of states that are accesible for a given system and energy. The corner stone of the theory is that the system is equally likely to be in any one of these states.We will call the number of given states g. They also introduce the entropy σ = ln g. We will also talk about systems in thermal contact. This will lead us into the second law of thermodynamic. The temperature is defined as 1 τ = ∂σ ∂u N, V where u is the internal energy. 3. The Boltzman factor: Imagine we have a small system in a hot bath, we find that the probability that it will be found in a given state is given by p(ε) ∝ e −ε/τ 4. The Planck Distribution: 5. Systems in both thermal and diffusive contact. This leads to a new concept called the chemical potential. The chemical potential we call μ which is defined as − μ τ = ∂σ ∂N U, V This leads us to something called the Gibbs factor. the probabilty of finding one particle having some energy in some state p(1, ε) ∝ e (μ−ε)/τ 1
Transcript of Physics 112: Thermodynamicsrogerg/education/112_formulas.pdf · 2008-05-06 · about systems in...
Physics 112: ThermodynamicsServed by Roger GriffithNutritional Facts:Serving size: 1 Semester (16 weeks)Servings per container: many problems and solutionsoffice hours for John Clarke Wed 3-4 349 Birge
Preamble for this course
Why do we need statistical mechanics?
• Quantum Mechanics: Particle in a box, Harmonic oscillator,H-atom. The SE gives us our solutions.|ψ|2 gives us the probability of finding a particle somewhere withsome speed. But this is notstatistics. But for a cm3 of air ~ 3×1019 molecules, this is a hopeless problem to solve the SE.But for a macroscopic system we must use a statistical approach. e.g to find the heat capacity orparamagnetic susceptibility. Statistically this turns out to yield exceedingly accurate results.
Road Map
1. Define states of the system, probability, binomial distribution, gaussian, and mean value theorem.
2. Entropy: The measure of the number of states that are accesible for a given system and energy. Thecorner stone of the theory is that the system is equally likely to be in any one of these states.Wewill call the number of given statesg. They also introduce the entropyσ = lng. We will also talkabout systems in thermal contact. This will lead us into the second law of thermodynamic. Thetemperature is defined as
1τ
=
(
∂σ∂u
)
N,V
whereu is the internal energy.
3. The Boltzman factor: Imagine we have a small system in a hotbath, we find that the probability thatit will be found in a given state is given by
p(ε) ∝ e−ε/τ
4. The Planck Distribution:
5. Systems in both thermal and diffusive contact. This leadsto a new concept called the chemicalpotential. The chemical potential we callµ which is defined as
−µτ
=
(
∂σ∂N
)
U,V
This leads us to something called the Gibbs factor. the probabilty of finding one particle havingsome energy in some state
p(1,ε) ∝ e(µ−ε)/τ
1
6. Ideal classical gas
7. Quantum Gases: They come in two flavors, Fermi gas and Bose gas. The Fermi gas is for fermionsand the Bose gas is for bosons. they have profoundly different behaviours.
8. Heat and Work: heat engines and refrigerators
2
Chapter 1
States of A Model System
1.1 Number Of States
1.1.0.1 Stationary Quantum State
The first thing we must talk about is the multiplicity or also known as degeneracy, we will call thisg of asystem with given energy which is equal to the number of quantum states with that energyε. e.g Particlein 3D box
nx ny nz
1 1 11 1 21 2 12 1 1
the first energy state is non degenerateg = 1 the second energy state is three-fold degenerateg = 3.Quantum states of a one-particle are called “orbitals”. ForNon interacting Spins
N= ↑ ↑ ↑ ↓ ↑ ↑ ↓ ↑
the total number of arrangements is given by 2N, the total magnetic moment is given byM = Nm for allspins that are↑ andM = −Nm for all spins that are↓. How many ways can we getM = (N−2)m (flipover any one spin). We will callg the number of ways in which we can get a particular value of amagnetic moment. Lets define the following
n↑ =
(
N2
+s
)
↑ n↓ =
(
N2−s
)
↓
whereN is even andM = (n↑−n↓)m= 2smwhere the 2s= spin excess= n↑−n↓. If we assume eachspin has a probabilityP↑ of being↑ and a probability ofP↓ of being↓ . we also know thatP↑ +P↓ = 1. Inthe presence of a magnetic field↑, P↑ > P↓. The probability of a configuartion with spin excess 2s is
PN/2+s↑ andPN/2−s
↓ . The probability of finding a magnetic momentM = (2s)m is given by the following
P(N,s) = g(N,s)PN/2+s↑ PN/2−s
↓
we need to find what the multiplicityg(N,s) is. In how many ways can we distributen↑ = N/2+sspinsoverN places. we can assume that they are distinguishable spins. So we can assume that the first spin canbe on any one of theN sides.
3
• 1st spin can be in any one of theN sides
• 2nd spin can be in any one of the (N−1) sides
• 3rd spin can be in any one of the(N−2) sides
• 4th spin can be in any one of the(N−3) sides
• nth spin can be in any one of (N−n↑ +1) sides
Thus the total number of the arrangements is
total = N(N−1)(N−2)...(N−n↑ +1)(N−n↑)
=N!
(N−n↑)!
We have to be careful of overcounting, we can address this in the following way. If we dont care aboutthe color of the spin or about which spin goes where, we have overcounted. The first of these spins
• 1st of spin up could have been any one of then↑ magnets
• 2nd of spin up could have been any one of then↑−1 magnets
• nth of spin up could have been any one of the 1 magnets
Thus we have overcounted byn↑(n↑−1)(n↑−2)...−1 = n↑!. Thus
g(N,s) =N!
(N−n↑)!n↑!
This is known as themultiplicity function ; it is the number of states having the same values. If wereplace
n↑ =N2
+s
yields
g(N,s) =N!
(
N2 −s
)
!(
N2 +s
)
!
This is the number of ways of achieving a spin excess of 2s. Hence
P(N,s) =N!
(N2 −s
)
!(N
2 +s)
!PN/2+s↑ PN/2−s
↓
For the moment lets assume that we dont apply any fieldB = 0thenP↑ = P↓ = 12, thus we find
P(N,s) =N!
(
N2 −s
)
!(
N2 +s
)
!
12N
If we have the following spins
N spins ↑ ↑ ↓ ↑ ↑ ↓the probability is given as
P(N,s) =g(N,s)
2N 2s= spin excess
where
g(N,s) =N!
(
N2 −s
)
!(
N2 +s
)
!
4
1.2 Gaussian Distribution
we first need to assume
N ≫ 1, N even |s| ≪ N P↑ = P↓ =12
we will use ln as the natural logrythim, thus
lng(N,s) = lnN!− ln
(
N2
+s
)
!− ln
(
N2−s
)
!
working on this expression yields(
N2
+s
)
! =
[
1,2,3...N2
](
N2
+1
)
, ...
(
N2
+s
)
=
(
N2
)
!
(
N2
+1
)
, ...
(
N2
+s
)
ln
(
N2
+s
)
! = ln
(
N2
)
! +s
∑k=1
ln
(
N2
+k
)
and now we need to do it for the other one(
N2−s
)
=1,2,3, ..
(
N2 −s
)(
N2 −s+1
)
, ...N2
N2 ....
(N2 −s+1
)
=
(N2
)
!
N/2......(
N2 −s+1
)
ln
(
N2−s
)
! = ln
(
N2
)
!0−s
∑k=1
ln
(
N2−k+1
)
hence
ln
(
N2
+s
)
! + ln
(
N2−s
)
! = 2ln
(
N2
)
! +s
∑k=1
ln
[
1+ 2kN
1− 2kN + 2
N
]
and since we know thats≪ N and alsok≪ N, we can expand the expression
s
∑k=1
[
ln
(
1+2kN
)
− ln
(
1− 2kN
+2N
)]
≈s
∑k=1
[
2kN
−(
−2kN
+2N
)]
≈ 4N
s
∑k=1
(
k− 12
)
when we add all these terms together and find that there ares terms, thus the sum is given by
4N
s
∑k=1
(
k− 12
)
=2s2
N
putting this all together yields
lng(N,s) = lnN!−2ln
(
N2
)
!− 2s2
N
5
thus
ln
[
g(N,s)[(
N2
)
!]2
N!
]
= −2s2
N
and
g(N,s) =N!
(
N2
)
!(
N2
)
!e−2s2/N
and to summarize
g(N,s) = g(N,0)e−2s2/N where g(N,0) =N!
(
N2
)
!(
N2
)
!
we also know that this function falls of as 1/ewhen
2s2
N= 1, s=
(
N2
)1/2 sN
=
(
12N
)1/2
thus, if N is very large, the peeking is exceedingly sharp. For largeN, thermodynamic quantities (heatcapacity,paramagnetic susceptibility are very well defined. To take an example, lets assume that
N = 1020 (1 cm3air)sN
≈ 10−11
How to evaluateN!? There is a useful approximation called Sterling’s approximation, it is defined as
for largeN N! ≈ (2πN)1/2NNe−N
we need to work with this until we make it look like the multiplicity g(N,s)
lnN! = N lnN−N+12
ln(2πN)
note that the error is less the 1% forN = 10 and it gets much better for higherN. Lets use this to evaluateg(N,0)
lng(N,0) = ln
[
N!(N
2
)
!(N
2
)
!
]
= N lnN−N+12
ln(2πN)−2
[
N2
lnN2− N
2+
12
ln(πN)
]
let us do
lng(N,0) = N ln2+12
ln(2πN)− 12
ln(πN)2 = N ln2+12
ln
(
2πN
)
or more compactly
lng(N,0) = ln
[
2N(
2πN
)1/2]
and finally
g(N,0) = 2N(
2πN
)1/2
6
1.2.1 Notes on The Gaussian
g(x,σ) = Ae−(x−x)2/2σ2
where
x = mean value
= 0 for P↑ = P↓6= 0 for P↑ 6= P↓
σ = standard deviation
whereσ is used for error analysis, this is a method of expressing theconfidence in your results.
±σ ≈ 68.3%
±2σ ≈ 98.4%
±3σ ≈ 99.7%
1.3 The Central Limit Theorem
The Gaussian distribution holds for any distribution provided that three things are true
• N is very large
• P(x) falls off sufficently rapidly asx→ ∞
• Events are statistically independent
1.4 Mean Values
Suppose that functionf (s) has a probability ofP(s) of occuring. Then the mean value is
〈 f 〉 = f = ∑s
f (s)P(s) where ∑s
p(s) = 1
e.g, the toss of a die
n =6
∑n=1
np(n) = (1+2+3+4+5+6)× 16
= 312
Example: Spin system
for P↑ = P↓ (i.e equally likely for spins to be↑ or ↓)thus
P(N,s) =g(N,s)
2N
and using the definition
〈 f 〉 = f = ∑s
f (s)g(N,s)
2N
7
if we letf (s) = s2
then〈s2〉 6= 〈s〉2
but
〈s2〉 = ∑s
s22N(
2πN
)1/2
e−2s2/N 12N
=
(
2πN
)1/2 Z ∞
−∞s2e−2s2/Nds
if we let
x2 =2s2
Ns=
(
N2
)1/2
x ds=
(
N2
)1/2
dx
and so we find
〈s2〉 =
(
2πN
)1/2 N2
(
N2
)1/2 Z ∞
−∞x2e−x2
dx=N4
now lets calculate the spin excess
〈(2s)2〉 = N mean square
〈(2s)2〉1/2 =√
N root mean square
to find the fractional rms spin excess〈(2s)2〉
N=
1√N
1.5 Summary
1. Multiplicity for N spins 1/2 with spin excess 2s is given by
g(N,s) =N!
(
N2 +s
)
!(
N2 −s
)
!
but forN ≫ 1, ands≪ N we can write
g(N, .s) = g(N,0)e−2s2/N
where
g(N,0) =N!
(N2
)
!(N
2
)
!= 2N
(
2πN
)1/2
and using Sterlings approximation
lnN! = N lnN−N+12
ln(2πN)
2. the Mean value〈 f 〉 = ∑
sf (s)P(s), where ∑
sP(s) = 1
and the spin system〈(2s)2〉
N=
1√N
8
1.6 Problems and Solutions
Problem # 1A penny is tossed 400 times. Use the Gaussian distribution tofind the probability of getting 215 heads.
A graphical representation of this is given as
Since we know that the Gaussian distribution is given as
g(N,s) =
(
2πN
)1/2
2Ne−2s2/N
And the probability is given as
P(N,s) =g(N,s)
2N
and since we know thatN = 400 s= x− x = 215−200= 15
we find that the probability of getting 215 heads is given as
P(400,15) =
(
2π(400)
)1/2
e−2(15)2/400≈ 1.3%
9
Problem # 2The Berkeley Campus has 2000 telephones (a very conservative estimate!). To guarantee access to the
outside world, 2000 telephone lines would be required, a rather extravagant number. Suppose that duringthe peak use hour of the day, each campus phone is used to make asingle two-minute call at a randomtime. Find the minimum number of telephone lines required sothat at most only 1% of callers fail to haveimmediate access to a line. (Hint: approximate the distribution by a Gaussian distribution, and use tablesor your computer.).
Method 1Since we know that in any 2 given minutes, the average number of people using a telephone line is
given byx = 〈x〉 = Np= 200030 . We find that this distribution can be modeled as a Gaussian distribution of
the form
The distribution plotted above is given by
p(x) =1√
2πσ2e−(x−x)2/2σ2
whereσ2 = Np(1− p)
wherep is the probability that a phone line will be used which is
p =130
N = 2000
so all we have to do is integrate this function to some value ofx that will yield 99%, thus
p(x < xl ) =1√
2πσ2
Z x
0e−(x−x)2/2σ2
dx
10
using IDL we are able to plot this function and integrate to get
x = 85.3≈ 86 lines
thus we would need a minimum of 86 lines so that at the most 1% ofthe people are not able to make acall.
N = 86 lines
Method 2If we say thatn↑ are the number of phone lines being used
n↑ =N2
+s n↓ =N2−s
and if we let
n↑ =N2
+s= N s=N2
and we also know that the probability is given as
P(N,s) = g(N,s)PN/2+s↑ PN/2−s
↓
where the probability of the number of lines being used is given by
P↑ =2 min60 min
2000 linesN lines
=2003N
and the other probability is given as
P↓ = 1−P↑ =3N−200
3N
The Gaussian approximation for the multiplicity is given as
g(N,s) = 2N(
2πN
)1/2
e−2s2/N
but since we know thats= N2 we get
g(N,N/2) = 2N(
2πN
)1/2
e−N/2
thus we find that the probability is given by
P(N,s) = 2N(
2πN
)1/2
e−2s2/N(
2003N
)N/2+s(3N−2003N
)N/2−s
(1.1)
given thats= N2 this simplifies into
P(N,N/2) = 2N(
2πN
)1/2
e−N/2(
2003N
)N
11
and since we know that the probability needs to be 0.01% we canjust write this as
2N(
2πN
)1/2
e−N/2(
2003N
)N
= 0.01
This can only be solved numerically, the numerical solutionto this is
N = 83.0125≈ 84 lines
Thus we know thatP(83,83/2) = 0.01
Method 3If we model this as a binomial distribution
p(n) =N!
(N−n)!n!pn(1− p)N−n
to find the minimum number of phone lines that will be needed sothat fewer than 1 % of the callers fail toget thru requires us to take a sum
p(n < nl) =N
∑nl+1
N!(N−nl )!nl !
pnl (1− p)N−nl < 0.01
wherenl is the minimum number of phone lines required to satisfy thiscriterion. This can easily be putinto Mathematica, but I dont have it so I will quote the solution as
nl = 86
Method 4If we look at the following function
p(x) =1√
2πσ2
Z x
0e−(x−x)2/2σ2
dx
We can see that this function resembles the error function
erf(x) =2√π
Z x
0e−u2
du
and the soultion forx of this function is given by the inverse error function
x = erf−1(erf(x))
but we know that erf(x) is simply the probability that we are looking for, which in our case we are lookingfor anx that will give us 99%. Thus we know that
u = erf−1(erf(x)) = erf−1(0.99) = 1.822
but we know that ouru is simplyx− x√
2σ= 1.822
12
since we know that
σ =√
Np(1− p) x =200030
thus we findx =
√2σ(1.822)+ x
thus we find that in order to get 99% we need
x = 86.681= 87 lines
we have just basically shown that this problem can be done 4 different ways all yielding similar result,it is just a matter of the approximation that we make.
Problem # 3The binomial distribution can be written in the form
P(N,n) =
(
Nn
)
pn(1− p)N−n(
Nn
)
=N!
n!(N−n)!
wherep is the probability that some single event occurs.Assume thatp≪ 1, andN ≫ n. Show thatP(N,n) can be written in the approximate form
P(N,n) =λne−λ
n!whereλ = Np. This is the well-known Poisson distribution (first derivedto study the death rate from
horse kicks in the French army!). SketchP(N,n) vs. n for λ < 1 andλ > 1.
Since we know that
P(N,n) =
(
Nn
)
pn(1− p)N−n
where the binomial coefficient is(
Nn
)
=N!
(N−n)!n!
or using Newton’s generalized binomial theorem(
Nn
)
=N(N−1)(N−2)...(N− (n−1))
n!
but since we know thatN ≫ n this is simply(
Nn
)
=Nn
n!
thus we can write our original expression as
P(N,n) =(Np)n
n!(1− p)N−n
we can see that that
(1− p)N−n =∞
∑k=0
(
N−nk
)
(−p)k
13
but we can see that(
N−nk
)
=(N−n)(N−n−1)(N−n−2)...(N−n− (k−1))..
k!=
Nk
k!
since we know thatN ≫ n. Thus we find
(1− p)N−n =∞
∑k=0
(−Np)k
k!= e−(Np)
thus we find
P(N,n) =(Np)n
n!e−(Np)
but if we letNp= λ then we recover what we were looking for
P(N,n) =λn
n!e−λ
and the Poisson curves are given by
We can see from the plots that the Poisson distribution is well defined for large values ofN, in actuallitythe Binomial distribution becomes the Poisson distribution for large value ofN.
Problem # 4 The Meaning of NeverIt has been said that “six monkeys, set to strum unintellegently on typewriters for for millions of years,
would be bound in time to write all the books in the British Museum.” This statement is nonsense, for itgives a misleading conclusion about very, very large numbers. Could all the monkeys in the world havetyped out a single specified book in the age of the universe?
Suppose that 1010 monkeys have been seated at typewriters throughout the age of the universe, 1018 s.This numner of monkeys is about three times greater then the present human population of the Earth. Wesuppose that a monkey can hit 10 typewriter keys per second. Atypewriter may have 44 keys; we acceptlowercase letters in place of capital letters. Asuuming that Shalespear’sHamlethas 105 characters, willthe monkeys hit uponHamlet?
14
a) Show that the probability that any given sequence of 105 characters typed at random will come out inthe correct sequence (the sequence ofHamlet) is of the order
(
144
)100000
= 10−164345
where we have used log1044= 1.64345.
Since we know that there are a total of 44 keys then we know thatthe probability that a key will land uponthe right place is given by
p(1) =144
but since we know that there are a total of 105 characters in total forHamlet we know that the totalprobability of this happening is given by
p(N) =
(
144
)N
but since we know thatN = 105 this is simply
p(N) =
(
144
)100000
but we know that
log10 p(N) = 100000log10
(
144
)
≈−164,345
thusp(N) = 10−164345
b) Show that the probability that amonkey−Hamletwill be typed in the age of the universe is approxi-mately 10−164316. The probability ofHamlet is therefore zero in any operational sense of an event,so that the original statement at the beggining of this problem will never occur in the total literaryproduction of the monkeys
Since we know that there is
Nmonkeys= 1010 tuniverse= 1018 s Nkeys/s=10 keys
s monkey
thus the total number of keys that these monkeys can type in the total age of the universe is
Nkeys= 1010 monkeys×1018 s× 10 keyss monkey
= 1029 keys
the probability that amonkey−Hamletwill be typed in the age of the universe is given by
p(monkey−Hamlet) = 1029×10−164345= 10−164316
Problem # 5 Coin Flips revisited
15
a) You flip 10 coins. What is the probability of obtaining exactly 5 heads and 5 tails?
We know thatN = 10 and that the spin excess 2s= n↑−n↓ = 0. We also know that the probability of thistwo-state system is given by
P(N,s) =g(N,s)
2N
we also know that
g(N,s) =N!
(N2 +s
)
!(N
2 −s)
!
thus
g(10,0) =10!5!5!
and the probability is given as
P(10,0) =10!5!5!
1210 = 0.25
b) You flip 1000 coins. What is the probability of obtaining exactly 500 heads and 500 tails?
We know thatN = 1000 and the spin excess is 2s= 0. Since we also know thats≪ N allows us to maketeh Sterling approximation
g(N,s) = g(N,0)e−2s2/N
where
g(N,0) = 2N(
2πN
)1/2
and so the probability is given by
P(N,s) =g(N,s)
2N =
√
2πN
thus we find
P(1000,0) =
√
21000×π
= 0.025
c) Why is the answer to a) more than b), yet we say the multiplicity function sharpens with increasingN.
The probability of getting exactly zero “spin excess”, i.e 50 % heads and 50 % tails decreases asNincreases. But the probability of being very close tos= 0 increases as 1/
√N. It is the fractional width of
the distribution which decreases.
16
Chapter 2
Entropy and Temperature
Entropy and Temperature (Roadmap)
1. Definitions
2. Equal a priori probaility
3. Two systems in thermal equlibrium: Most probable configuration (MPC)
4. Thermal equilibrium: Entropy and Temperature
• Definition of Entropy
• Definition of temperature
• Justification of MPC approximation
5. Laws of thermodynamics,0th,2nd,3rd
1. Some definitions
2.1 A Closed System
The energy, number of particles, external parameters( Electric fields and Magnetic fields are fixed)
2.1.1 A quantum state is accesible
if its properties ( energy, number of particles) satisfy thespecifications of a system. e.g lets take a cubicbox: for 1 particle the energy of the state is identified by three quantum number,nx,ny, andnz.
• One particle would remain in its state indefenitely
Now we add more particles- ecah one of these is in a state specified by these quantum numbers (nx,ny,nz)with a total fixed energy
• Collisions cause transitons of the particles among quantumstates
• The total energy of a particlemust be conserved
Those states with that given fixed energy are accessible states.
17
2.2 Macroscopic Properties
The macroscopic propertie that we measure (heat capacity, suceptibility,...) are time averages of thesesystems. However, in statistical mechanics we consider an ensemble average. The ensemble containsgidentical systems. One for each accesible state, whereg is the multiplicity. We then average over theensemble to obtain a given macroscopic properties.
Example
we have 4 spin (1/2) with 2s= 2, the number of accesible states
g(N,s) = g(4,1) =N!
n↑!n↓!=
4!3!1!
= 4
Thus there are 4 possible states, so lets now take four systems, where each one is accesible to system
↓ ↑ ↑ ↑
↑ ↓ ↑ ↑
↑ ↑ ↓ ↑
↑ ↑ ↑ ↓
This is our ensemble, lets suppose that the probability of finding the system i an accesible statek withspin excess 2 is
2 = p(k)
where
∑k
p(k) = 1
Lets suppose that we have a parameterx that takes the valuex(k) when the system is i the statek, thiscould be the magnetic suceptibility or the heat capacity. Then the expectation value ofx is
〈x〉 = ∑k
p(k)x(k) ensemble average
It is reasonable to suppose that thetime average is the same as the ensemble average, this is known asErgodic Hypothesis, but there is no proof.
2.3 Postulate of Equal Apriori Probability
a priori - before testing, before hand, inate..
"A closed system is equaly likely to be in any one of its accesible states"
We have to bear in mind that this is a postulate and cannot be derived. i.e Newton’s three laws, orMaxwell’s equations. All we can do is to build a theory based on this postulate and then we can com-pare the results with the experiments. If we get predictionsthat are correct, than we ca say our postulateis justified “a posteori” (after testing). Thus
p(k) =1g
18
and thus
〈x〉 =1g∑
k
x(k)
Example
Lets suppose that we have the four magnetic spins all with different electric dipole moments. Lets letxbe the electric dipole momentd,2d,3d, and 4d. For the four systems we have the following net electricdipole moments
↓ ↑ ↑ ↑
↑ ↓ ↑ ↑
↑ ↑ ↓ ↑
↑ ↑ ↑ ↓
−d+2d+3d+4d = 8d
2−2d+3d+4d = 6d
d+2d−3d+4d = 4d
d+2d+3d−4d = 2d
So the expectation value is
〈d〉 =∑x(k)
g=
8d+6d+4d+2d4
= 5d
2.4 Two Systems in Thermal Contact: Most Probable Configuration(MPC)
As an example, consider spin system. Lets suppose that thereis a magnetic fieldB
N1,u1,s1 system 1 N2,u2,s2 system 2
whereS is the spin excess, this is a closed system with a total energyET . When we put this two togetherwe get a new system
N1,u′1 N2,u′2
This is also a closed system in thermal contact. But we have the constraint that
u1+u2 = u′1+u′2 = u
s1+s2 = s′1+s′2 = S
19
the total energyu = −2smB= −2(s1+s2)mB= −2(s′1+s′2)mB
We also need to assume thatN1 < N2, and thus we have the following constraint
|s1| ≤N1
2or −N1 ≤ 2w≤ N1
lets write down the multiplicityg1(N1,s1) system 1
g2(N2,s2) system 2
so the multiplicity of the combined system is given by the following
g(N,s) =N1/2
∑s1=−N1/2
g1(N1,s1)g2(N2,s2) =N1/2
∑s1=−N1/2
g1(N1,s1)g2(N2,s−s1)
We have removedS2. These products have a maximum value which is to be determined for some value ofS1, this is what is called the most probable configuartion (MPC). For large values ofN we will show thatwe need to keep only the most probable configuration of the sum. i.e
g1(N1, s1)g2(N2,s− s1)
lets write this in terms of a Gaussian, we saw that
g1(N1,s1)g2(N2,s−s1) = g1(N1,0)g2(N2,0)e−2s21/N1e−2(s−s1)
2/N2
Lets plots1 versusg(N,s).The maximum value correspond to where these two curves intersect. This isknown as the most probable configuration.
s1 = s1 +δs2 = s2−δs21 = s1
2+2s1+δ2
s22 = s2
2−2s2+δ2
Thus
g1(N1, s1+δ)g2(N2, s2−δ) = g1(0)g2(0)exp
[
−2s12
N1− 4s1δ
N1− 2s1δ
N1− 2s1
2
N1+
4s2δN2
− 2δ2
N2
]
= (g1g2)exp
[
−2δ2
N1− 2δ2
N2
]
In the general case we know that
g1(N1,s1)g2(N2,s−s1) = g1(N1,0)g2(N2,0)e−[2s21/N1−2(s−s1)
2/N2]
As we increases1 this exponetial grows, thus we get a maximum value ˆs1. What we need to do is figureout this value, where we get the following result.
g1(N1, s1+δ)g2(N2, s2−δ) = (g1g2)exp
[
−2δ2
N1− 2δ2
N2
]
20
Lets plug some numbers in to evaluate this, lets take a cm3 of a solid, the number of atoms in this solid is
N1 = N2 = 1022
lets suppose thatδ is
δ = 1012 sN1
= 10−10
Thus2s2
N= 200
i.e g1g2
(g1g2)max= e−400≈ 10−173
we can get a bit of insight in the following way, we can say thatthis configuration will never occur! Thesystem works its way through all possible values ofg1 andg2, so for this particular configuration to reachthe configurationg1 andg2 once, the system will be in(g1g2)max 10173 times. So if we suppose that anyspin flips every 10−12 seconds. Thus the number of configurations per second will be
1022
10−12 = 1034/sec
so the time taken to get this deviation of
δN1
= 10−10∼ 10173
1034 ∼ 10139sec∼ 10132yrs
where we know that the age of the universe is approximately 1018 seconds, thus we can conclude that thisconfiguartion will never occur. This is a very important result and tells us why we will only consider themost probable configuartion MPC.
2.4.1 Find maximum value ofg1 and g2
We know thatg1 andg2 can be written in the form
g1g2 = g1(0)g2(0)exp
[
−2s2
N1− 2(s−s1)
2
N2
]
if we take the log of both sides we get
lng1g2 = ln[g1(0)g2(0)− 2s2
N1− 2(s−s1)
2
N2
if we take the derivative we find
∂∂s1
[lng1g2] = −4s1
N1+
4(s−s1)
N2= 0 ⇒ s1
N1=
s−s1
N2=
s2
N2
thus the fractional spin excess are equal, if we take the second derivative we find
∂2
∂s21
[lng1g2] = − 4N1
− 4N2
< 0
21
so ats1
N1=
s2
N2=
sN
which is the fractional excess of the combined system, thus we have
s1
N1=
s2
N2=
s− s1
N2=
s− s1
N−N1
which becomes
s1
(
1N1
+1
N−N1
)
=s
N−N1
thus
s1 =
[
N1−N1+N1
N1(N−N1)
]
=s
N−N1
which simplifies intos1
N1=
sN
using this result we finds1
2
N1+
s22
N2=
s1s1
N1+
s2s2
N2=
sN
(s1+ s2) =s2
N
thus we can conclude that
(g1g2)max= g1(0)g2(0)e−2s2/N
2.5 Thermal Equilibrium: Entropy and Temperature
We have the expression for the multiplicity of the two systems given as
g(N,u) = ∑u1≤u
g1(N1,u1)g2(N2,u2)
if we assume that we keep only the most probable configuration, we can drop the sum to get
g(N,u) = g1(N1,u1)g2(N2,u2) MPC
if we take the derivative we find
dg(N,u) =
(
∂g1
∂u1
)
g2du1+
(
∂g2
∂u2
)
g1du2 = 0
wheredu1 +du2 = 0
if we now take this expression and devide
1g1
(
∂g1
∂u1
)
N1
=1g2
(
∂g2
∂u2
)
N2
this relates the derivative of the first system to the second system, hence
(
∂ lng1
∂u1
)
N1
=
(
∂ lng2
∂u2
)
N2
22
this is the condition for thermal equilibrium. This is wherewe will introduce entropy
σ(N,u) = lng(N,u) Entropy
the entropy is the measure of the degree of randomness in a system. It is the natural log of the multiplicity.Most people define entropy as
S= kB lng
as a final remark, Entropy is additive
σ(N,u) = ln[g1(N1,u1)g2(N2,u2)
= lng1(N1,u1)+ lng2(N2,u2)
= σ1(N1,u1)+σ2(N2,u2)
We can define the temperature as1τ
=
(
∂σ∂u
)
N
thus in equilibriumτ1 = τ2
conventionallyτ is written asτ = kBT
we will relateτ to the Kelvin scale of absolute temperature.
2.5.1 Jusitification for keeping only the MPC
If we assume thatg = ∑
u1
g1(N1,u1)g2(N2,u2) = (g1g2)max
we also assumed in our discussion of entropy that
σ = lng(N,u) = ln(g1g2)max= σ1 +σ2
the correct expression is really
σ = ln[
∑g1(N1,u1)g2(N,u−u1)]
6= σ1+σ2
Why can we neglect all the other terms? How valid is our approximation? We have the following expres-sion
g1(N1,s1)g2(N2,s2) = (g1g2)maxexp
[
−2δ2
N1− 2δ2
N2
]
δ = s1− s1
lets suppose that we write
N1 = N2 =N2
we get
g(N,s) = ∑δ
g1(N/2, s1+δ)g2(N/2, s1−δ)
= (g1g2)max
Z ∞
−∞e−8δ2/Ndδ
23
so in our analysis we have just ignored the gaussian, the integral is
Z ∞
−∞e−8δ/Ndδ =
√
N8
Z ∞
−∞e−x2
dx=
√
Nπ8
hence
g(N,s) = (g1g2)max
√
Nπ8
next we can use Sterling’s approximation
g(N,0) = 2N(
2πN
)1/2
N → N2
we get the following
g(N/2,0) =
(
2N/2
√
4πN
)2√Nπ8
= 2N 4πN
√
Nπ8
finally we need to form the entropy
σ(N,s) = lng(N,s) = N ln2+ ln
(
4πN
)
+12
ln
(
Nπ8
)
we can rewrite this as follows
σ(N,s) = N ln2− lnN+12
lnN+12
ln(π/8)+ ln(4/π)
= N ln2− 12
lnN
if we take a big numberN = 1022
N ln2∼ 1022 lnN ∼ 2.3log101022 ∼ 50
thus the only term we really care about is
σ(N,s) ≈ N ln2
this term dominates all others for largeN, thus we can conclude that only the most probable configu-ration contributes to the entropy and the temperature.
We have been talking about thermal equilibrium of two systems, the system was dominated by theMPC. We have defined the Entropy and the temperature of the system. We will concentrate on is the lawsof thermodynamics
• Zeroth Law- two systems in thermal equilibrium with a third system then they are all in thermalequilibrium
τ1 = τ3
τ2 = τ3
τ1 = τ2
24
• The First Law- Heat is a form of energy
• The Second Law- The Entropy of a closed system will either remain constant or increases when aconstraint is removed.
If we remember the two system multiplicity
g = ∑u
g1(u1)g2(u−u1)
somewhere in this sum is the term that corresponds to the starting point. What we will call that as
g(u) = gi1(u1(t = 0))gi
2(u−u1(t = 0))+ ∑u1 6=u1(t=0)
g1(u1)g2(u−u1)
what we have is two terms, where the first term is the entropy
σ f = ln(g) = ln(gi1gi
2)+ ln(
∑g1g2)
whereσi = ln(gi
1gi2)
we know that the final value of the entropy is
σ f ≥ σi
If we can imagine having two systems whereτ1 > τ2 and thatδu is removed from system and added tosystem 2. What happens to the entropy when we do this
∆σ =
(
δσδu1
)
N1
(−δu)+
(
δσδu2
)
N2
(δu)
∆σ = δu
(
1τ2
− 1τ1
)
≥ 0
whenτ1 > τ2 thenδu > 0
• The Third Law of Thermodynamics- The entropy of a system approaches a constant value as thetemperature goes to zero.
This will follow from our statistical approach if temperature is an increasing function of energy(
δTδu
)
N> 0
again we will have two systems where we haveδu removed from system 2
δu1 = δu δu2 = −δu τ1 = τ2 = τ
again we have to go back to our law of increase of entropy, whathappens to∆σ1 and∆σ2, this are thevalues att = 0. We can do this by using a Taylor expansion
σ1 = σ0+
(
∂σ1
∂u1
)
δuu0
+12
(
∂2σ1
∂u21
)
u0
(δu)2+ ..
σ2 = σ0+
(
∂σ2
∂u2
)
(−δu)u0
+12
(
∂2σ2
∂u22
)
u0
(−δu)2+ ..
25
and
∆σ1 =δuτ
+(δu)2
2∂
∂u
(
∂σ1
∂u1
)
u0
∆σ2 = −δuτ
+(δu)2
2∂∂u
(
∂σ2
∂u2
)
u0
and finally∆σ = ∆σ1 +∆σ2
is∂
∂u
(
∂σ∂u
)
=∂
∂u
(
1τ
)
= − 1τ2
∂τ∂u
thus
∆σ =(δu)2
2
(
− 1τ2
∂τ1
∂u− 1
τ2
∂τ2
∂u
)
= −(δu)2
2τ2
(
∂τ1
∂u+
∂τ2
∂u
)
< 0
the above function is evaluated atu0, thus we have just shown that
∂τ∂u
≥ 0
As τ is lowered then the energy decreases, thus at absolute zeroτ = 0 then the system is in the lowestenergy state and it is in a non degenerate state.
σ → ln(1) = 0 σ → σ0
2.6 Summary
• Equal a priori probability -
• Ergodic hypothesis - This is simply that the time average = ensemble average. The ensemble averageis simply
〈x〉 = ∑s
X(s)p(s)
which is a weighted average over all possible configurations
• Multiplicity function of combined system
g(N,u) = ∑u1<u
g1(N1,u1)g2(N2,u−u1)
for largeN the sum was dominated by one term, and this term was called themost probable config-uration MPC.
• From the multiplicity function we defined the entropy as
σ = ln(g(u)) S= kBσ
this was an extensive property, andsince we use the log then this made this an additive quantity.
26
• We then defined the temperature as
1τ
=
(
∂σ∂u
)
Nτ = kBT
• With these two notions we are able to talk about thermal equilibrium
τ1 = τ2 σ is maximized
we can consider temperature to be an intensive property thatincreases with energy.
• Finally we also talked about the law of increase of entropy. Two system is thermal equilibrium
∆σ ≥ 0
• The laws of thermodynamics
0th τ1 = τ3 τ2 = τ3 τ1 = τ2
1st = heat= energy
2nd = ∆σ ≥ 0
3rd = limτ→0
σ = σ0
2.7 Problems and Solutions
Problem # 1 Entropy and TemperatureSupposeg(U) = CU3N/2, whereC is a constant andN is the number of particles.
a) Show that
U =32
Nτ
Since we knowg(U) = CU3N/2
we can take the natural log of both sides
lng(U) = ln(
CU3N/2)
= ln(C)+3N2
ln(U)
Taking the partial derivative on both sides with respect toU yields
∂∂U
lng(U) =∂
∂U
[
ln(C)+3N2
ln(U)
]
=3N2
∂∂U
ln(U)
we know that the left hand term is simply the entropy
σ = lng(U)
27
thus∂σ∂U
=3N2
1U
thus
U =3N2
∂U∂σ
but we know that the temperature is defined as
1τ
=
(
∂σ∂U
)
N,V
thus we have just showed that
U =3N2
τ
b) Show that(
∂2σ∂U2
)
N< 0
show that it is negative. This form ofg(U) actually applies to an ideal gas.
We can simply take a double derivative ofg(U)
∂∂U
(
∂∂U
lng(U)
)
=∂
∂U
(
3N2
∂∂U
ln(U)
)
this becomes∂2σ∂U2 =
∂∂U
(
3N2
1U
)
= − 3N2U2
since we know thatN can only be positive and the energy can only be positive we have just shown
(
∂2σ∂U2
)
N< 0
Problem # 2 ParamagnetismFind the equilibrium value at temperatureτ of the fractional magnetization
MNm
=2〈s〉N
of the system ofN spins each of magnetic momentm in a magnetic fieldB. The spin excess is 2s. Takethe entropy as the logarithim of the multiplicityg(N,s) as given
σ(s) ≃ logg(N,0)− 2s2
N(2.1)
for |s| ≪ N. Hint : Show that in this approximation
σ(U) = σ0−U2
2m2B2N(2.2)
28
with σ0 = logg(N,0). Further, show that 1/τ = −U/m2B2N, whereU denotes〈U〉, the thermal averageenergy.
Since we know thatσ0 = logg(N,0) U = −2smB
thus
s= − U2mB
thus Equation 1 becomes Equation 2, i.e
σ(U) = σ0−U2
2m2B2N
if we take a differential on both sides with respect toU we find
∂σ(U)
∂U= − U
m2B2N
but we know that1τ
=∂σ∂U
= − Um2B2N
so we find that the average thermal energy is given by
〈U〉= −m2B2Nτ
this is related to the spin excess and the magnetic moment by
〈s〉 = − 〈U〉2mB
=mBN
2τ
we find that the equilibrium value at temperatureτ of the fractional magnetization is given as
MNm
=mBτ
Problem # 3 Addition of entropy for two spin systemsGiven two systems ofN1≃N2 = 1022 spins with multiplicity functionsg1(N1,s1) andg2(N2,s−s1), the
productg1g2 as a function ofs1 is relatively sharply peaked ats1 = s1. Fors1 = s1+1012, the productg1g2
is reduced by 10−174 from its peak value. Use the Gaussian approximation to the multiplicity function;the form
g1(N1, s1+δ)g2(N2, s2−δ) = (g1g2)maxe
(
− 2δ2N1
− 2δ2N2
)
(2.3)
may be useful.
a) Computeg1g2/(g1g2)max for s1 = s1+1011 ands= 0.
29
Fors1 = s1+δ we know thatδ = 1011
δ = 1011 N1 ≃ N2 = 1022 s1+s2 = s ⇒ s1 = −s2
We can see that Equation 3 can be written as
g1(N1,s1)g2(N2,s−s1)
(g1g2)max= e−4δ2/N1 ≈ 0.018
Fors= 0 we know thatδ = 0 thus we know that
g1g2
(g1g2)max= 1
b) For s= 1020, by what factor must you multiply(g1g2)max to make it equal to∑s1g1(N1,s1)g2(N2,s−
s1); give the factor to the nearest order of magnitude.
Method 1
since we know that
g1(N1,s1)g2(N2,s2) = g1(N1, s1+δ)g2(N2, s2−δ) = (g1g2)maxe
(
− 2δ2N1
− 2δ2N2
)
sinces1 = s1+δ s2 = s2−δ
thus
∑s1
g1(N1,s1)g2(N2,s−s1) = (g1g2)max∑δ
e
(
− 2δ2N1
− 2δ2N2
)
= (g1g2)max×Q
whereQ is the factor we need to multiply(g1g2)max so that it is equal to∑s1g1(N1,s1)g2(N2,s−s1). So
we can just turn this sum into an integral
Q =
Z ∞
−∞e
(
− 2δ2N1
− 2δ2N2
)
dδ =
Z ∞
−∞e
(
− 4δ2N1
)
dδ
this is simply a Gaussian integral with a solution
Q =
√
πN1
4= 8.86×1010≈ 1011
Method 2
We know thatQ× (g1g2)max= ∑g1(N1,s1)g2(N2,s−s1) (2.4)
but we know that we can write the sum as
∑g1(N1,s1)g2(N2,s−s1) = g(N,s)
since we know thatN = N1+N2 = 2N1
30
thus we findg(2N1,s) = g(2N1,0)e−2s2/2N1 = g(2N,0)e−s2/N1
we also know that
(g1g2)max= g1(N1,0)g2(N2,0)e−2s2/2N1 = g1(N1,0)g2(N2,0)e−s2/N1
thus Equation 4 can be written as
Q =∑g1(N1,s1)g2(N2,s−s1)
(g1g2)max=
g(2N1,0)
g1(N1,0)g2(N2,0)(2.5)
but we know that
g(N,0) =
√
2πN
2N
and Equation 5 becomes
Q =
√
22πN1
22N1
√
2πN1
√
2πN1
22N1
=
√
πN1
4= 8.86×1010≈ 1011
c) How large is the fractional error in the entropy when you ignore this factor?
We know that the fractional error is defined as
∆σσ
=σ′−σ
σ′
whereσ′ = ln(g1g2)maxQ σ = ln(g1g2)max
thus we find
∆σσ
=lnQ
ln(g1g2)maxQ=
lnQlnQ+ ln(g1g2)max
=lnQ
lnQ+ lng1(N1,0)g2(N2,0)− s2
N1
but we know that(g1g2)max= g1(N1,0)g2(N2,0)e−s2/N1
and we also know that
g1(N1,0) =
√
2πN1
2N1 g1(N1,0) =
√
2πN2
2N2
since we know thatN1 = N2 = 1022, plugging this into some calculating machine we find
∆σσ
≈ 1.83×10−21
this shows us that the extra factor does not affect the entropy very much.
31
Chapter 3
Boltzman Distribution & Helmholtz FreeEnergy
• Resevoirs and systems
• Boltzman factor
• Partition Function
• Example of a paramagnetic system
• Pressure
• Thermodynamic identity
• Helmholtz free energy
• Classical Ideal Gas
• Equipartition Theorem
3.1 Resevoir & Systems
We have a resevoir in a large system, what we assume is that they are in thermal contact at some temper-atureτ and also that the energy of the system isεl ≪ u0. We must consider an entire ensemble of suchsystems and then we must take the average of these. If we keepN fixed then we call this the cononicalensemble. We must write down the multiplicity of the combinbed system
g(u0) = gR(u0− εl )gs(εl)
what we are specifying withl is the label of a particular quantum state of a system. Thus there is only oneway for it to be in that state which means
gs(εl) = 1
thusg(u0) = gR(u0− εl)
and since this total system is equally likely to be in any of ofthese states thus the probability is
pl = AgR(u0− εl )
32
and since this is a probability
∑l
pl = 1
again we have an approximation which implies that we will have to do a Taylor expansion, but we willfind that these multiplicity functions will begin to get out of hand when Taylor expanded. We will insteadTaylor expand the entropy
σ = ln[gR(u0− εl )] ≈ ln[gR(u0)]−(
εl∂ ln[gR]
∂u
)
u0
+ ...
= ln
[
gR(u0− εl )
gR(u0)
]
= −εl
τ
If we raise both sides to the exponetial
gR(u0− εl) = gr(u0)e−εl/τ
thus the probability is given as
pl ∝ e−εl /τ This is known as the Boltzman factor
Last time we talked about resevoirs, where we made the assumption that the energy in the system is muchsmaller than the energy in the resevoir
εl ≪U0
wherel is to denote the state of the system. What we are looking for isthe ensemble average〈X〉
〈X〉 = ∑l
X(l)p(l)
we need to find the multiplicity function of the combined system
g(U0) = gR(U0− εl)gs(l)
wheregs(l) = 1
by definition. Thus the probability we are looking for is proportional tog(U0), and to do this we mustTaylor expand to find
g(U0− εl ) = Ae−εl /τ
where the probability isp(l) ∝ e−εl/τ
which is known as the Boltzman factor.If we look at some simple model systems, i.e SHO
εn =
(
n+12
)
~ω
we can see that the higher the energy the lower the probability of finding this system in this state
pn+1
pn=
e−(n+3/2)~ω/τ
e−(n+1/2)~ω/τ = e−~ω/τ
33
and we can see
τ → 0pn+1
pn→ 0
τ → pn+1
pn→ 1
and we also know that
∑l
pl = ∑l
Ae−εl/τ ⇒ A
(
∑l
e−εl /τ
)−1
where
pl =e−εl/τ
∑l e−εl/τ
where the denominator is known as the partition function
Z ≡ ∑l
e−εl /τ
this is useful if you start with a microscopic model where we know how to calculate the microstateenergies. From these microstate energies we can use the partition function to get the thermodynamicproperties of this system. If we define the energy states thenwe can use the partition function as
Zn = ∑n
gs(εn)e−εn/τ
wheregs(εn) is how many states haveε = εn. Lets find the average energy given the partition function
U ≡ 〈ε〉 = ε
where
〈ε〉 = ∑l
εl p(l) =1Z ∑
l
εle−εl/τ
where we often useβ = 1/τ, using this we get
〈ε〉 = ∑l
εl p(l) =1Z ∑
l
εl e−εl β
and the trick we can use∂
∂βe−εl β = −εl e
−εl β
and now we can write
〈ε〉 =1Z
(
−∂Z∂β
)
= −∂ lnZ∂β
to put this back in terms ofτ we just simply need to use the chain rule
∂Z∂τ
=∂Z∂β
∂β∂τ
=∂Z∂β
(
− 1τ2
)
thus
〈ε〉 =1Z
τ2∂Z∂τ
= τ2∂ lnZ∂τ
We will need to calculate what the scale of the energy fluctuations are, we will use a method that wasused in quantum mechanic.
34
3.2 Energy Fluctuations
〈(∆ε)2〉 mean squared energy fluctuation
where(ε−〈ε〉)
and
〈(∆ε)2〉 = 〈ε2−2ε〈ε〉+ 〈ε〉2〉= 〈ε2〉−2〈ε〉〈ε〉+ 〈ε〉2
= 〈ε2〉−〈ε〉2
we can use the same trick we have done before
〈ε2〉 =1Z ∑
l
ε2l e−βεl
where the trick is∂2
∂β2e−βεl = ε2l e−βεl
thus
〈ε2〉 =1Z
(
∂2Z∂β2
)
∂∂β
(
1Z
∂Z∂β
)
= − 1Z2
(
∂Z∂β
)2
+1Z
∂2Z∂β2
we can write this as
− 1Z2
(
∂Z∂β
)2
+1Z
∂2Z∂β2 = −
[
1Z
∂Z∂β
]2
+ 〈ε2〉 = −〈ε〉2+ 〈ε2〉
to finish up
〈(∆ε)2〉 =∂
∂β
(
∂ lnZ∂β
)
=∂2 lnZ
∂β2
3.3 Paramagnetic System
We haveN number of atoms in a volume where each atom has a spin of 1/2 anda magnetic momentm. Weapply a magnetic fieldH and we need to find〈m(τ)〉. The system consist of a single spin and the resevoirhas all the other spins with respect toτ. We will also neglect all the spin-spin interactions.
〈m(τ)〉=1Z ∑
l
ml e−βεl
what are the states? there are only two states, 1 toward the field and one opposing the field
m↑ = +m ε↑ = −mB spin up
m↓ = −m ε↓ = mB spin down
and we also know thatB = µH, now we find the partition function to be
Z = e−βε↑ +e−βε↓ = eβmB+e−βmB
35
so the expectation value is now
〈m(τ)〉=meβmB−me−βmB
eβmB+e−βmB= m
sinh(βmB)
cos(βmB)= mtanh(βmB)
thus〈m(τ)〉 = mtanh(βmB)
but we are looking for the moment for allN spins
〈M(τ)〉= N〈m(τ)〉 = Nmtanh
(
βmτ
)
we can also find the suceptibility to be
χ =d〈M〉dB
=Nm2
τsech2
(
mBτ
)
if we look at the limits of the hyporbolic tangent function
tanh(x) =ex−e−x
ex +e−x
thus for smallx we find
limx→0
tanh(x) =(1+x)− (1−x)
1+x+1−x=
2x2
= x
and for largex we findlimx→∞
tanh(x) ≈ 1
In the limit thatmB/τ ≪ 1 or high temperature
〈M(τ)〉 ≈ Nm
(
Bmτ
)
=Nm2B
τ
and
χ =Nm2
τCurie’s law
in the limit of low temperature we find〈M(τ)〉 = Nm
andχ = 0
another thing we are interested in is to find the average energy U = 〈ε〉
〈ε〉 = −〈M(τ)〉B = −NmBtanh
(
mBτ
)
thus the heat capacity is defined as
CV =
(
∂U∂T
)
VkB
(
∂U∂τ
)
=
(
∂U∂T
)
36
where∂U∂τ
= −NmBsech2(
mBτ
)(
−mBτ2
)
thus the heat capacity is
CV =kBNm2B2
τ2 sech2(
mBτ
)
lets look at what the heat capacity does in the extremes. At high temperatures
sech(x) =2
ex +e−x =2
1+x+1−x= 1
and
CV =Nm2B2
τ2
and at low temperatures
sech(x) =2ex = 2e−x
in this case the heat capacity is roughly
CV =Nm2B2
τ2
(
2e−mB/τ)2
=4Nm2B2
τ2 e−2mB/τ
if we look at this graphically we can see that there is a maximum that is known as the Schotkki anomoly.If you have interaction we get another phenomenom
χ → 1τ− τC
whereτC is known as the Curie temperature. This form tells us that when the temperature is lower thanthe Curie temperature forces a negative susceptibility andwe have a feromagnetic material and not para-magnetic.
At high temperatures the magnetic spins are all equally divided in up states and down states. At lowtemperatures all of the spins are in the up (high energy) state and the magnetic energy is higher than thethermal energy. If all the spins are aligned with he field it will lower the energy and if all the spins areanti-aligned that raises the energy.
3.4 Calculating the average energy using the partition function
we can write the average energy as
〈ε〉− 1Z
∂Z∂β
= −∂ lnZ∂β
= τ2∂ lnZ∂Z
where we know that the partition function is given as
Z = emβB +e−mBβ = 2cosh(mβB)
and thus∂Z∂β
= 2mBsinh(mβB)
so we have
〈ε〉 = −1Z
∂Z∂β
= −2mBsinh(mβB)
2cosh(mβB)= −mBtanh(mβB)
which is the exact result we have already obtained.
37
3.4.1 Negative Temperature
We know that the energy is always an increasing function of temperature. We will need to consider a twolevel quantum system. We can write the ratio of the number of particles in one energy state to the numberof particles in the other energy state
N2
N1=
e−E2/τ
e−E1/τ = e−(E2−E1)/τ > 1
whereE2−E1 is assumed positive, for this to be true means thatτ < 0. The negative temperature meansthat we have a higher population in the high energy state thanin the lower energy state. This is known asa non-equilibrium system. This happens in lasers because weneed more atoms in the high energy statethan in the lower energy state. To get laser action we must have all of the atoms in the upper state makingtransitions to the lower energy state that. This is also knowas population inversion. We also know thatnegative temperatures require an upper bound on the energy,where this is given byNmB. This also meansthat the multiplicity is bounded. This would not be true in the case of a gas, because as you raise thetemperature you also raise the energy. There is no negative temperatures for systems that have kineticenergy or that the value of the multiplicity increases without bound. Negative temperatures do have moreenergy than positive energies. If we have two systems, one with negative temperature and one with positivetemperature that are in thermal contact we find thet heat flowsfrom theτ < 0 toτ >0.
3.5 Pressure
We define quantities such asσ andτ in terms of thermodynamic varaibles. But the pressurep must bedefined so as to be consistent with Newton’s laws.
Newton’s laws
If we have a cylinder with a gas at an initial positionx and we expand the volume by an amount∆x with aconstant area and a constant force pushing down on the gas. Ifthe piston moves up by a distance∆x thenwe know that the work doneon the gas is given by
∆W = −F∆x = −F∆VA
= −p∆V
we also find this to be equal to∆W = ∆U
which is the increase of the energy of the system, and hence the pressure is given by
p =∂U∂V
we need to go through a slightly different approach to get to the thermodynamic relationships.
3.5.1 Thermodynamics
Consider a system in a quantum statel . If we letV →V +∆V (quasistatically), this means that the systemremains in thermal equilibrium. Imagine a box with distinctenergy levels that are populated by the atomsor molecules of the gas. If we now expand the system we know that the energy levels will get closer
38
together. Because the system remained in thermal equilibrium than on the average the population of thisenergy levels will not change. This is know as theadiabatical approximationwhere the state of thesystem remains the same and hence the entropy remains fixed. We can now wrtite the energy as
εl (V) → εl(V +∆V)
in the same state asl , we can now do a Taylor expansion to find
εl (V +∆V) = ε(V)+dV
(
∂εl
∂V
)
σwhere the lower case sigma means that this process is at constant entropy. The ensemble average is givenby
U(V +∆V) = U(V)+dV
(
∂U∂V
)
σ= U(V)− pdV
and thus
p = −(
∂U∂V
)
σwe need to know how to get pressure in terms of entropy. What weknow is that we can write
σ = σ(U,V,N) whereN is constant
If we let
0 = dσ =
(
∂σ∂U
)
V,N(dU)σ +
(
∂σ∂V
)
U,N(dV)σ
and now if we do
0 =
(
∂σ∂U
)
V,N
(
∂U∂V
)
σ+
(
∂σ∂V
)
U,N
where we know that1τ
=
(
∂σ∂U
)
V,N− p =
(
∂U∂V
)
σand we finally find
pτ
=
(
∂σ∂V
)
U,N
3.6 Thermodynamic Identity
To simplify this we will writeσ = σ(U,V) and we will keepN fixed, thus
dσ =
(
∂σ∂U
)
VdU +
(
∂σ∂V
)
UdV
=1τ
dU +pτ
dV
we can now write this asτdσ = dU + pdV
thus we can write the change in the energy as
dU = τdσ− pdV
whereτdσ is the heat added to the system and−pdV is the work done on the system.
39
3.7 Helmholtz Free Energy
We have used the free energy in the thermodynamic identity as
dU(σ,V) = τdσ− pdV
this does not always have to be the case, imagine a systme at constant temperature. We could write thisinstead as
dU = d(τσ)−σdτ− pdV
whered(U − τσ) = −σdτ− pdV
anddF(τ,V) = −σdτ− pdV
so the quantityF(τ,V) = U − τσ
this is known as Helmholtz free energy. If we take a derivative of this we get the following
dF(τ,V) =
(
∂F∂τ
)
Vdτ+
(
∂F∂V
)
τdV
and hence
σ = −(
∂F∂τ
)
Vp = −
(
∂F∂V
)
τ
whereF is the appropriate energy for isothermal systems (fixedτ). If we look at the definition for thepressure
p = −(
∂F∂V
)
τ= −
(
∂(U − τσ)
∂V
)
τ= −
(
∂U∂V
)
τ+ τ(
∂σ∂V
)
τ
where the first term is the energy part and the second term is the entropy part. The first tirm tends todominate for solids, wheras if you had a gas, this term would be really small. On the other hand thesecond term dominates for a gas. There is one final thing we canget from this algebra, this is what iscalled “the Maxwell’s relations”
3.8 The Maxwell Relations
From
σ = −(
∂F∂τ
)
Vp = −
(
∂F∂V
)
τ
If we differentiate(
∂σ∂V
)
τ= −
(
∂2F∂V∂τ
) (
∂p∂τ
)
V= −
(
∂2F∂τ∂V
)
thus we can see(
∂σ∂V
)
τ=
(
∂p∂τ
)
V
40
Why is the Helmholtz free energy interesting? We know that the free energy is a minimum for a system inthermal equilibrium with a resevoir.
F = U − τσ
imagine a resevoir wheredFs = dUs− τσs fixed temperature
we also have the following
1τ
=
(
∂σ∂U
)
VdUs = τσs constant volume
if we substitute this into the above equation we find
dFs = 0
is this a maximum or a minimum? Lets consider the total energyof the system
U = UR+Us
σ = σR+σs = σR(U −Us)+σs(Us)
if we assumeUs≪UR then we can simply do a Taylor expansion
σ = σR−Us
(
∂σR
∂UR
)
V+σs(Us)
we can recall that1τ
=
(
∂σR
∂UR
)
V
putting this in gives
σ = σR(U)− 1τ
(Us−σsτ)
we can see that
σ = σR−Fs
τthis is because that we are assuming thatσR(U) is a fixed quantity. We know that in thermal equilibriumthat the entropy is a maximum, therefor, if we consider a small deviation fromσ
∂σ = ∂σR−∂Fs
τ= −∂Fs
τ
thus forτ > 0 then ifσ is a maximum means thatFs is a minimum. On the other hand the energy of thesystem corresponds to a minimum.
Example: Paramagetism
Lets find the Helmholtz free energyF and the average magnetizationM of N spins atτ ≫ mB (hightemperature). We can write the free energy as
F(s,τ,B) = U(s,τ,B)−σ(s,τ,B)τ
41
where we know thatU(s,τ,B) = −2smB
and alsog(N,s) = g(N,0)e−2s2/N
this is the Gaussian approximation in the high temperature limit. Lets put all this together
F(s,τ,B) = −2smB− τ lng(N,0)+2s2τN
finally, in thermal equilibrium we can take the derivative(
∂F∂s
)
τ,B= −2mB+
4sτN
= 0
thus
2s=mBN
τwhere
M = 2sm=m2BN
τthis is the same thing we found in the high temperature limit before tanh→ 1.
3.9 Helmholtz free energy and the Partition Function
If we let
F = U − τσ σ = −(
∂F∂τ
)
V
thus
F = U + τ(
∂F∂τ
)
V
and we also know
τ2 ∂∂τ
(
Fτ
)
V= −F + τ
(
∂F∂τ
)
V
and we find
F = U + τ2 ∂∂τ
(
Fτ
)
V+F
and finally
U = −τ2 ∂∂τ
(
Fτ
)
V
but recall that
U = τ2∂ lnZ∂τ
and thus we findFτ
= − lnZ
42
this is a nice way to relate the microscopic propertiesZ to the macroscopic propertiesF . We can also writethis expression as
Z = e−F/τ
if we remember the probability
p(εl ) =e−εl/τ
Z= e(F−εl )/τ
thus we are able to calculate the Boltzman factor if we know the free energy of the system without everknowing what the partition function is.
3.10 One Particle in a Box
For simplicity we assume that our box is a cube, with sidesl . There is one particle with massM. We allknow how to write down the Schrodinger Equation and we can findthe energies as
εm =~
2
2m
(πL
)2(n2
x +n2y +n2
z) n2 = n2x +n2
y +n2z n2
x +n2y +n2
z ≥ 1
we can now write the partition function
Z1 = ∑nx
∑ny
∑nz
e−~
2π2
2mL2τ(n2
x+n2y+n2
z)
since we know that the temperature is much greater than te energy spacing allows us to turn this sum intoan integral. This is know as the classic limit approximation. If we let
α2 =~
2π2
2mL2τ
thus we can writeZ1 =
Z ∞
0dnx
Z ∞
0dny
Z ∞
0dnze
−α2(n2x+n2
y+n2z)
but this can also be written as
Z1 =
[
Z ∞
0e−α2n2
xdnx
]3
the solution is
Z1 =π3/2
823/2M3/2L3τ3/2
~3π3 = VnQ
wherenQ is known as the quantum concentration and defined as
nQ =
(
Mτ2π~2
)3/2
43
3.11 Average energy
U = τ2(
∂ lnZ1
∂τ
)
V
now
lnZ1 =32
lnτ+ τ independent terms
thus(
∂ lnZ1
∂τ
)
V=
32
1τ
or
U = τ2 32τ
=32
τ =32
kBT
and the specific heat is given by
CV =32
kB
this allows us to measure the mean square velocity
12
M〈v2〉 =32
kBT
thus
〈v2〉 =3kBT
MWe want to know what the meaning of the concentration is? Recall the Debroigle wavelength
λD =hp≈ h
M(3kBT/M)1/2≈ h
(3Mτ)1/2
and
λ3D ≈ h3
(3Mτ)3/2≈ 1
nQ
thus
nQ ∼ 1
λ3D
this means that the concentration corresponds to one particle in a volume ofλ3D. This allows us to under-
stand degeneracy in gases. If 2 particles are separated by a distance much greater thenλD then their wavefunctions do not overlap and they are in the “classical” regime, and the results do not depend whether theyare bosons or fermions. On the other hand if they are separated by less than the Debriogle wavelengththen their wave functions overlap and they are in the “quantum” regime. Now the results depend cruciallyon whether they are bosons or fermions.
3.11.1 Example: N particles in a box
Imagine that we have a number of particlesN2 at standard pressure and temperature STP. we know that
n≈ 6×1023 molecules/mole22.4×103 cm3/mole
≈ 3×1019 molecules/cm3
44
what is the quantum concentration? we know that the mass is given by
M ≈ (1.67×10−27 kg)×28≈ 4×10−26 kg
and
n≈(
4×1026×1.4×10−23×3002π(10−34)2
)
≈ 1026 molecules/cm3
thus we are very strongly in the classical limit. The seperation is given by
δ = 102×λ
3.12 Equipartition Theorem
we know that the energy for a single atom is given by
U =32
τ
and that “each” “square” term in momentum and position coordinates has a mean energyτ/2 (kBT/2) inthe classical limit. For an ideal gas, the hamiltonian is given by
H =p2
x
2m+
p2y
2m+
p2z
2m
and
U =32
τ
Lets imagine that we only have one dimension. Thus the Hamiltonian is given by
H =p2
2m+
12
kx2
and we have an energy given byU = τ
Proof
Lets imagine a one dimensional gas with a Hamiltonian
H =p2
2m= Bp2
i
which is just a general expression for a quadratic term. The energy is given by
εi =
R ∞−∞ εie−βεi dpiR ∞
0 e−βεi dpi=
− ∂∂β
R ∞−∞ e−βεi dpi
R ∞−∞ e−βεi dpi
this is simplified as
εi = − ∂∂β
ln
[
Z ∞
−∞εie
−βεi dpi
]
45
and nowZ ∞
−∞εie
−βεi dpi =
Z ∞
−∞εie
−βBp2i dpi
if we lety = β1/2pi
we getZ ∞
−∞εie
−βεi dpi =1√
β
Z ∞
−∞εie
−By2dy
but we know that
ln
[
1√
β
Z ∞
−∞εie
−By2dy
]
= −12
lnβ+ terms independent ofβ
thus the average energy is
εi = − ∂∂β
(
−12
lnβ)
=1
2β=
τ2
now we would like to apply this to the idea that all resistor generate Johnson noise or Nyquist noise.
3.13 Johnson Noise or Nyuist noise
Imagine that you have some resistor with a value ofR and if we were to measure the noise across aVN,so as a function of time we would get something in which its average is zero, but the mean square valueis not zero. We could then measure the current noise by connecting a wire at the end of the circuit. Wewill once again see a signal that fluctuates. This provides anabsolute limitation on how accurately youcan measure something. We will apply the equipartition function to see how this works. We will calculatethe power spectrum which is also known as the spectral density. This is usually given asSν( f ) but in ourcase it is convenient to write it asSV(ω) the following relationships are
SV(ω)dω = SV( f )d f
where
Sv(ω) = SV( f )d fdω
we want to look at a narrow window that has a 1 Hz bandwidth, thus this is the mean square voltage in a1 Hz bandwidth, as you will see, as we scan the frequency the value of Swill vary. If we can imagine acircuit with a resistor, a conductor and a capacitor. The quality factor for this circuit is given by
Q =ω0LR
ω20 =
1LC
we can also writeQ as
Q =ω0
∆ω∆ω =
RL
Now using the equipartition we can see that there will be two square terms, thus we get the following
12C〈V2
N〉 =12
kbT 〈V2N〉 =
kBTC
and12
L〈I2N〉 =
12
kBT 〈I2N〉 =
kBTL
46
all frequencies add up to contribute to the square terms. Thetask is to go from the mean square term tothe spectral density. So Lets write down themean square value for the current source
〈I2N〉 =
kBTL
=
Z ∞
0
SV(ω)dω|Z(ω)|2
this takes a voltage squared and devide by the impedence squared. We all remember that we can write
Z(ω) = R+ iωL+1
iωC
and we also rember that1C
= ω20L
thus we find that
Z(ω) = R+iLω
(ω−ω20)
2
but
|Z(ω)|2 = L2[
R2
L2 +(ω2−ω2
0)2
ω2
]
∆ω2 =R2
L2
this allows us to write
〈I2N〉 =
Z ∞
0
SV(ω)
L2
dω
∆ω2+(ω2−ω2
0)2
ω2
this function is sharply peaked aboutω0 for highQ. If we take the limit we find
SV(ω) → SV(ω0)
if we look at(ω2−ω2
0)2
ω2 =((ω−ω0)(ω+ω0))
2
ω2
if we let ω → ω0 we find(ω2−ω2
0)2
ω2 = 4(ω−ω0)2
given this we get the following result
kBTL
=SV(ω0)
4L2
Z ∞
0
dω(ω−ω0)2+R2/4L2 x = ω−ω0 = dω a = R/2L
as you can see this now looks more tractable, thus we can now write this as
SV(ω0)
4L2
Z ∞
0
dxx2 +a2
where we all know thatZ ∞
0
dxx2 +a2 =
πa
thus we findkBTL
=SV(ω0)π
4L2(R/2L)=
SV(ω0)
2LRπ
47
thus we find that the spectral
SV(ω0) =2π
kBTR
this answer does not depend on eitherL or C, thus this is true for all values ofL andC. This is true for allvalues ofω. This is avery good example of what is called “white noise”. Thus we have the final result
SV(ω) =2π
kBTR SV( f ) = 4kbTR
this is the famous results calledNyquist noise or Johnson noise. The current noise is given by
SI( f ) =4kBT
R
The Nyquist noise sets a fundamental limit on the smallest voltage or current that one can measure. Forexample if we have some experiment that can be modeled with a resistor connected to an amplifier thenthe smallest value of the voltage will always be limited by the Nyquist noise. What we will see is thatthe magnitude scales linear with the absolute temperature.Thus to make more accurate measurements wemust decrease the tempearture. This enables us to measure lower voltages or anything else we might wantto measure. If we lower the temperature enough then we will reach the quantum limit of the noise, whichwill take over whenh f ≥ kBT. As a final remark because of the linear dependence with temperature thanthe Nyquist noise is used as an absolute thermometer.
3.14 Summary
The Boltzman factor
p(εl ) =e−εl/τ
Zis the probability using the Boltzman factor. We then introduced the partition function
Z = ∑s
e−εs/τ
we then intruduced the concept of pressure, the pressure is given by
p = −(
∂U∂V
)
σ= τ(
∂σ∂V
)
U
where we have the identity given bydU = τdσ− pdV
this allows us to write the Helmholtz free energy
F = U −στ
this is only for isolated thermal systems. We also defined theentropy as
σ = −(
dFdτ
)
V
48
and
p = −(
dFdV
)
τthe Maxwell relations are given by
(
∂σ∂V
)
τ=
(
dpdτ
)
V
where the free energy isF = −τ lnZ
and the probability is given byp(εl ) = e(F−εl )/τ
we then went to a particle in a box, where
Z1 = nQV nQ =
(
Mτ2π~2
)3/2
where the energy of one particle is given by
U =32
τ
we then move on to the Equipartition theorem
U =τ2
per quadratic term. Finally we talked about the Nyquist noise
SV( f ) = 4kBTR SI ( f ) =4kBT
R
3.15 Problems and Solutions
Problem # 1A dilute solution of macromolecules at temperatureT is placed in a centrifuge rotating with angular
velocity ω. The mass of each molecule ism. The equivalent centrifugal force on each particle in therotating frame of reference ismω2r, wherer is the radial distance from the axis of rotation.
Find how therelative density of moleculesρ(r) varies withr.
The probability that a system will be in a specific quantum state s of energyεs is proportional to theBoltzman factor. The ratio of the probability that the system is an a quantum state 1 at energyε1 to theprobability that the system is in quatum state 2 at energyε2 is just the ratio of the two multiplicities
P(ε1)
P(ε2)=
gR(U0− ε1)
gR(U0− ε2)(3.1)
the multiplicity is the number of states having the same value s. This is a direct consequence of what iscalled the fundemental assumption. If the resevoir is very large, than the multiplicities are very large. Wecan write Equation 1 in terms of the entropy of the resevoir
σ = lngR gR = eσ
49
thus Equation 1 now becomesP(ε1)
P(ε2)=
e(σR(U0−ε1))
e(σR(U0−ε2))(3.2)
we can define∆σR = σR(U0− ε1)−σR(U0− ε2) (3.3)
this allows us to write Equation 2 asP(ε1)
P(ε2)= e∆σR
since we know thatU0 ≫ ε allows us to Taylor expand the entropy (Equation 3). The Taylor expansion isdefined as
f (x0+a) = f (x0)+a
(
d fdx
)
x=x0
+12!
a2(
d2 fdx2
)
x=x0
+ ...
thus we find
σR(U0− ε1) = σR(U0)− ε1(∂σR/∂U)V,N + ...
= σR(U0)− ε1/τσR(U0− ε2) = σR(U0)− ε2(∂σR/∂U)V,N + ...
= σR(U0)− ε2/τ
thus we find∆σR = −(ε1− ε2)/τ
and we finally find thatP(ε1)
P(ε2)= e−(ε1−ε2)/τ
thus the ratio of the number of particles in a particular quantu state 1 to the number in a particular state 2is given by the ratios of their probabilities
N1
N2=
P(ε1)
P(ε2)= e−(ε1−ε2)/τ
we know that the density is defined as
ρ =mV
N
thus we know thatρ1 =
mV
N1 ρ2 =mV
N2
and the relative density is given by
ρ =ρ1
ρ2=
N1
N2= e−(ε1−ε2)/τ
which is simply the Boltzman factor. Since we also know that the centrifugal force is defined as
Fc = mω2r
and the energy is given by
ε = −Z
F ·dr = −12
ω2r2
50
thus we know that
ρ =ρ(r)ρ(0)
= eε1/τ
thus we find the relative density to be given as
ρ = e12τ mω2r2
a plot is given as
Problem # 2Show that for a system in thermal contact with a resevoirσ = lnZ+U/τ. This is a very useful result,
as you will see in (b) below.Consider a crystalline solid containingN atoms whose nuclei have spin one, so hat each nucleus can
be in one of the three statesms = 1,0,−1. We assume that the electric charge distribution in the nucleusis ellipsoidal, so that the energy of a nucleus depends on itsspin orientation with respect to the internalelectric field of the crystal. Thus, the energy isε for ms = 1 or -1, and zero forms = 0.
a) Find an expression for the nuclear contribution to the internal energy,U .
First we need to show that
σ = lnZ+Uτ
we can do this by knowing that the free energy is defined
F = −τ lnZ F = U −στ
putting these two expressio equal to each other yields
−τ lnZ = U −στ
51
which reduces to
σ = lnZ+Uτ
Next we need to find the average energy of the system (ensemble average)this is given by
U = 〈εs〉 = ε
where
〈ε〉 = ∑l
εl p(l) =1Z ∑
l
εle−εl/τ
where we often useβ = 1/τ, using this we get
〈ε〉 = ∑l
εl p(l) =1Z ∑
l
εl e−εl β
and the trick we can use∂
∂βe−εl β = −εl e
−εl β
and now we can write
〈ε〉 =1Z
(
−∂Z∂β
)
= −∂ lnZ∂β
to put this back in terms ofτ we just simply need to use the chain rule
∂Z∂τ
=∂Z∂β
∂β∂τ
=∂Z∂β
(
− 1τ2
)
thus
〈ε〉 =1Z
τ2∂Z∂τ
= τ2∂ lnZ∂τ
Thus the energy is given by
U = 〈ε〉 = τ2∂ lnZ∂τ
(3.4)
This results allows us to write the energy of a system by just knowing what the partition function is.
b) Write down the partition functionZ, of a single nucleus, and hence the partition functionZ for Nnuclei. Hence, find the entropy of the solid.
We know that the partition function is defined as
Z = ∑s
e−εs/τ
and we know thems = ±1 εs = ε and forms = 0 εs = 0. Thus we find thet the partition function for thissystem is given by
ZN = (1+2e−ε/τ)N
and the entropy of the solid is given by
σ = N lnZ+Uτ
52
using Equation 4 and the partition function we find the entropy to be given as
σ = N ln(1+2e−ε/τ)+ τ∂ ln(1+2e−ε/τ)
∂τ
now we just need to find∂ ln(ZN)
∂τ=
NZ
∂Z∂τ
and we find∂∂τ
(1+2e−ε/τ) = 2∂∂τ
(e−ε/τ)
letting
u = −ετ
∂u∂τ
=ετ2
thus we find
τ∂ ln(1+2e−ε/τ)
∂τ=
2Nετ
e−ε/τ
1+2e−ε/τ
thus we find the total entropy to be given as
σ = N ln(1+2e−ε/τ)+2Nε
τe−ε/τ
1+2e−ε/τ
a plot is given by
Where we plotτ/ε instead ofε/τ.
c) By directly counting the number of accessible states, calculate the entropy asτ → 0 andτ → ∞. Showthat your expression in (b) reduces to these values.
Since we know that the probability goes as
p =1g
thus the multiplicity for 1 particle goes as
g =1p
53
and the multiplicity forN particles is given by
g(N) =
(
1p
)N
and asτ → ∞ we know all the states are equally probable thus
g(N) = 3N p =13
and thus at this limit the entropy isσ = lng = N ln3
at the other extreme whenτ → 0 we know that all particles will all be in the ground state andthus theprobability is
p = 1 g = 1N
and the entropy isσ = lng = N ln1 = 0
now to show this with our previous solutions we just need to take the limits of our function to find
In the limit thatτ goes to 0
limτ→0
ln(1+2e−ε/τ)+2ετ
e−ε/τ
1+2e−ε/τ ≈ 0
this makes sense because we know if the temperature goes to 0 then the entropy goes 0.
In the limit thatτ → ∞ we find
limτ→∞
ln(1+2e−ε/τ)+2ετ
e−ε/τ
1+2e−ε/τ ≈ ln3
which also makes sense, as the temperature goes to infinity the entropy converges to a constant number.
d) Calculate the heat capacityCV of the crystal, and find its temperature dependence for highτ. Make asketch ofCV vs τ.
We know that the heat capacity is defined as
CV =
(
∂U∂τ
)
V
at constant volume. We know that the energy is given as
U = τ2∂ lnZ∂τ
= 2Nεe−ε/τ
1+2e−ε/τ
we need to find∂U∂τ
= 2Nε∂∂τ
(
e−ε/τ
1+2e−ε/τ
)
54
Using the quotient rule∂∂x
(
fg
)
=g f ′− f ′g
g2
if we say
f = e−ε/τ f ′ =ετ2e−ε/τ
and
g = 1+2e−ε/τ g′ =2ετ2 e−ε/τ
thus∂U∂τ
= 2Nε
[
(1+2e−ε/τ)(ε/τ2)e−ε/τ −e−ε/τ(2ε/τ2)e−ε/τ
(2e−ε/τ +1)2
]
this can be simplified by factoring out a term
∂U∂τ
=2Nε2
τ2 e−ε/τ
[
1+2e−ε/τ −2e−ε/τ
(2e−ε/τ +1)2
]
=2Nε2
τ2
e−ε/τ
(1+2e−ε/τ)2
thus the heat capacity for this system is given by
CV =2Nε2
τ2
e−ε/τ
(1+2e−ε/τ)2
in order to sketch this function we can write it in a more suggestive form and use IDL to plot it, if we let
x =ετ
we can write the heat capacity as
CV = 2Nx2 e−x
(1+2e−x)2
and the plot is given as
55
Problem # 3 Free energy of a two state system
a) Find an expression for the free energy as a function ofτ of a system with two states, one at energy 0and one at energyε.
Since we know that the partition function gives
Z = 1+e−ε/τ
and the free energy is defined asF = −τ lnZ
thus the free energy is given by
F = −τ ln(1+e−ε/τ)
b) From the free energy, find expressions for the energy and entropy of the system.
The entropy is defined as
σ = lnZ+Uτ
but we know the energy is defined as
U = τ2∂ lnZ∂τ
56
thus
σ = lnZ+ τ∂ lnZ
∂τwhere
τ∂ lnZ
∂τ=
ετ
(
e−ε/τ
1+e−ε/τ
)
thus the entropy is defined as
σ = ln(1+e−ε/τ)+ετ
(
e−ε/τ
1+e−ε/τ
)
with a plot given by
Problem # 4 Magnetic susceptibility
a) Use the partition function to find an exact expression for themagnetizationM and the susceptibilityχ ≡ dM/dB as a function of temperature and magnetic field. The result for the magnetization isM = nmtanh(mB/τ), as derived in (46) by another method. Heren is the particle concentration.
We know that the average magnetic moment is given as
〈m(τ)〉=1Z ∑
l
ml e−βεl
57
what are the states? there are only two states, 1 toward the field and one opposing the field
m↑ = +m ε↑ = −mB spin up
m↓ = −m ε↓ = mB spin down
and we also know thatB = µH, now we find the partition function to be
Z = e−βε↑ +e−βε↓ = eβmB+e−βmB
so the expectation value is now
〈m(τ)〉=meβmB−me−βmB
eβmB+e−βmB= m
sinh(βmB)
cos(βmB)= mtanh(βmB)
thus〈m(τ)〉 = mtanh(βmB)
but we are looking for the moment for allN spins
〈M(τ)〉= N〈m(τ)〉= Nmtanh
(
Bmτ
)
but we know that
M(τ) =N〈m(τ)〉
V= nmtanh
(
Bmτ
)
n =NV
thus
M(τ) = nmtanh
(
Bmτ
)
we can also find the suceptibility to be
χ =d〈M〉dB
=nm2
τsech2
(
mBτ
)
thus
χ =nm2
τsech2
(
mBτ
)
b) Find the free energy and express the result as a function onlyof τ and the parameterx≡ M/nm.
We know that the free energy is given asF = −τ lnZ
and we know that the partition function is given as
Z = eβmB+e−βmB
thus the free energy is given as
F = −τ ln(eβmB+e−βmB) = −τ ln2cosh
(
mBτ
)
58
and to put it in terms ofx≡ M/Nmwe simply do
x =M
Nm= tanh
(
mBτ
)
thusmBτ
= tanh−1(x)
we also know that
cosh2(x)−sinh2(x) = 1 cosh(x) =1
√
1− tanh2(x)
and finally we find that
F = −τ ln(2√
1−x2)
c) Show that the susceptibility isχ = nm2/τ in the limit mB≪ τ.
we know that for largex we findlimx→∞
tanh(x) ≈ 1
In the limit thatmB/τ ≪ 1 or high temperature
〈M(τ)〉 ≈ nm
(
Bmτ
)
=nm2B
τ
and
χ =nm2
τwe plotM/nmvs mB/τ
59
Problem # 6
N molecules ofN2 are condensed onto a surface of areaL×L. The molecules do not interact with eachother, so each molecule is completely free to move translationally in 2 dimensions (along the surface) andto vibrate, with vibrational frequencyν = 1012 Hz and non-degenerate energy levelsε = s~ω with sbeing any positive integer (ignoring zero point energy). Assume theN2 are distinguishable and thatL islarge so that
~2
2M
(πL
)2≪ ~ω
the following relationship will be useful
∞
∑n=0
aqn =a
1−qfor |q| < 1
this is known as a geometric series.
a) Write down an exact expression for the energyU as a function of temperatureτ. You do not need toderive this expression but please justify either in words orfrom simple mathematical expressionswhere it comes from.
We know that the translational energy given by a 1-D ideal gasis given by
U =12
Nτ
but since we have a 2-D system (2 degrees of freedom) we find thetotal translational energy to be
Utran = Nτ
This comes from solving for the energy levels and wavefunctions for particles in anL× L box. ForNdistinguishable particles, the partition function is
ZN =
[
∞
∑s=0
(e−~ω/τ)s
]N
and using the expression for the geometric series we find
ZN =
[
1
1−e−~ω/τ
]N
we know that the energy is given as
Uvib = τ2∂ lnZ∂τ
= τ2N(1−e−~ω/τ)
(1−e−~ω)2 e−~ω/τ ~ωτ2 =
N~ωe~ω/τ −1
thus the total energy of this system is given by
Utot = Nτ+N~ω
e~ω/τ −1
60
b) Calculate the exact specific heatCV as a function of temperatureτ and plot this showing units andvalues on the axes.
We know that the specific heat is defined as
CV =
(
∂U∂τ
)
V= N+
N~ω(e~ω/τ −1)2
e~ω/τ ~ωτ2
thus the specific heat is given by
CV = N
[
1+
(
~ωτ
)2 e~ω/τ
(e~ω/τ −1)2
]
Problem # 7ConsiderN particles, each of which can be found in one of four states: the ground state has energy
ε1 = 0 and is non-degenerate. The first excited state has energyε2 = ε and is triply degenerate. The systemis in thermal contact with a resevoir at temperatureτ.
a) Find the partition function for the system.
We know that the partition function of this system is given by
Z = ∑s
e−εs/τ = 1+3e−ε/τ = 1+3e−βε
and forN particles we find
ZN =[
1+3e−βε]N
b) FindU,F, andσ as a function ofτ.
We know that the energy is defined as
U = τ2∂ lnZ∂τ
= −1Z
∂Z∂β
we know whatZ is now we need to find
∂Z∂β
= N[1+3e−βε]N−1[−3εe−βε]
which can simplify into
U =3Nε
eε/τ +3
To find the free energy we know that
F = −τ lnZN = −τN[1+3e−ε/τ]
and the entropy is given as
σ = lnZN +Uτ
= N ln[1+3e−ε/τ]+3Nε
τ
[
1
eε/τ +3
]
61
c) Find the specific heatCV as a function ofτ and calculate its low and highτ limits.
We know that the heat capacity is given as
CV =
(
∂U∂τ
)
V
where we can see that∂U∂τ
=3Nε
(eε/τ +3)2eε/τ ε
τ2 = 3N(ε
τ
)2 eε/τ
(eε/τ +3)2
thus the heat capacity is given as
CV = 3N(ε
τ
)2 eε/τ
(eε/τ +3)2
for high temperature limitτ ≫ ε we can see that
eε/τ → 1+ετ≈ 1
thus
CV =3N16
(ετ
)2
for the low temperature limit we can see thatτ ≪ ε gives
eε/τ ≫ 3
thus
CV = 3N(ε
τ
)2e−ε/τ
Problem # 8Consider again the system described in Problem 7. For each particle, the ground state is a spin singlet
with s= 0 andε = 0 and the first excited state is a spin triplet withs= 1. With the applied magnetic fieldB= 0, the threes= 1 states have equal energyε as described above. (The energysplittingε in B= 0 is dueto effects other than magnetiuc field, specifically what are called crystal field splittings). In a magneticfield, thes= 0 state still has energy 0 (ms = 0), but thes= 1 states split into three states(ms = ±1,0) ,with energiesεs = ε+2µBB,ε,ε−2µBB for ms = −1,−,+1 respectively whereµB is the Bohr magneton.
a) Calculate the partition function as a function ofB andτ.
We know that the partition function for this system is given by
ZN = ∑s
e−εs/τ = [1+e−(ε+2µBB)/τ +e−(ε−2µBB)/τ +e−ε/τ]N
b) Calculate the expectation value of the momentM of the system as a function of fieldB (at fixedtemperatureτ). Hint M = N〈ms〉
62
Since we know that the expectation of the moment is given as
M = N〈ms〉 = N1Z
∞
∑s=0
mse−εsτ = N
1Z
[e−(ε−2µBB)/τ −e−(ε+2µBB)/τ]
this becomes
M = Ne−(ε−2µBB)/τ −e−(ε+2µBB)/τ
ZN=
Ne−ε/τ
ZN[e2µBB/τ −e−2µBB/τ] =
Ne−ε/τ2sinh(2µBB/τ)ZN
Problem # 9Consider a system that has 5 possible states, with two energylevels. The lowest energy level is doubly
degenerate (two states) with energyε1 = 0. The next energy level has threefold degeneracy (three states)with energyε2. It is in thermal contact with a resevior at temperatureτ.
a) What is the probability ( as a function of temperature) that the system has energy 0?
We know that the probability that the system will have energyε = 0 is given by
p(ε = 0) =e−ε/τ
Z=
2Z
where the partition function is given by
Z = ∑s
e−εs/τ = 2+3e−ε2/τ
thus the probability is given as
p(0) =2
2+3e−ε2/τ
b) Calculate the energyU and the entropyσ (as a fucntion ofτ).
We know that the energy is given by
U = −1Z
∂Z∂β
where∂Z∂β
= −3ε2e−βε2
thus the energy is given as
U =3ε2e−ε2/τ
2+3e−ε2/τ
And we know that the entropy is defined as
σ = lnZ+Uτ
= ln[2+3e−ε2/τ]+3ε2
τ
[
e−ε2/τ
2+3e−ε2/τ
]
thus we find
σ = ln[2+3e−ε2/τ]+3ε2
τ
[
e−ε2/τ
2+3e−ε2/τ
]
63
c) Calculate the energy and the entropy asτ goes to 0 andτ goes to infinity. Briefly discuss/explain youranswers.
We can see that asτ goes to 0 we finde−ε2/τ → 0
thusU → 0 σ → ln2
asτ goes to 0, the system goes into the ground state which haveε = 0⇒U → 0. Asτ → ∞
e−ε/τ → 1− ετ
thus we find
U → 3ε2(1− ετ)
2+3(1− ετ)
=3ε2(1− ε/τ)5−3(ε/τ)
the entropy is found using the same method.
Problem # 10
a) Prove the Maxwell relation(
∂σ∂V
)
τ=
(
∂p∂τ
)
V
for all systems. To do this, please start from
dσ(U,V) =
(
∂σ∂U
)
VdU +
(
∂σ∂V
)
UdV dU= τdσ− pdV
and the thermodynamic definitionF = U − τσ
calculatedF. From this, calculatep andσ in terms of derivatives ofF (show your calculations ofthe derivatives explicitely, i.e where they come from, don’t just state the result) and from this, provethe Maxwell relation.
We know that
dσ(U,V) =
(
∂σ∂U
)
VdU +
(
∂σ∂V
)
UdV =
dUτ
+pτ
dV
we also know thatdU = τdσ− pdV
andF = U − τσ
wheredF = dU− τdσ−σdτ = τdσ− pdV− τdσ−σdτ = −pdV−σdτ
thus(
∂F∂V
)
τ= −p
(
∂F∂τ
)
V= −σ
64
if we take the cross-derivatives(
∂∂τ
(p)
) (
∂∂τ
(σ)
)
which yields∂2F
∂τ∂V= −
(
∂p∂τ
)
V
∂2F∂v∂τ
= −(
∂σ∂V
)
τ
the order of the secon derivatives do not matter and thus we find
(
∂p∂τ
)
V=
(
∂σ∂V
)
τ
b) There is one important variable which is held constant in theabove Maxwell relation- what is it?
It is N that is held constant.
c) Show that the Maxwell relation is true for the ideal monatomic gas by explicitly calculatingF,σ andpand their derivatives for the ideal monatomic gas. Start from Z for the ideal monatomic gas, whichyou do not need to re-derive
Z =1
N!(nQV)N nQ =
(
Mτ2π~2
)3/2
We know that the free energy is given by
F = −τ lnZ = −τ ln1
N!(nQV)N = −τ[− lnN! +N lnnQ+N lnV]
we also know that the entropy is defined as
σ = −(
∂F∂τ
)
V,N= τN
∂ lnnQ
∂τ= τN
1nq
∂nQ
∂τ= τN
32
nQ
τ=
32
N
thus
σ = [− lnN! +N lnnQ +N lnV]+32
N
and for the pressure we find
p = −(
∂F∂V
)
τ,N= τN
∂ lnV∂V
=τNV
and we can finally show that(
∂σ∂V
)
τ,N= N
∂ lnV∂V
=NV
and(
∂p∂τ
)
V,N=
NV
65
Chapter 4
Thermal Radiation and The PlanckDistribution
Roadmap
1. Electromagnetic Modes
2. Density of Nodes
3. The Rayleigh-Jeans classical theorem of electro-magnetic radiation
4. Planck theory of electro-magnetic theorem (Quantum theory)
5. Absorption and Emission of EM radiation
6. Phonos in a solid.
4.1 Electromagnetic Modes
If we have some cavity in thermal equilibrium then it is filledwith EM radiation, if it is in thermal equi-librium then the modes only depend on the temperature. We need to calculate this which gives rise to thePlanck distribution. The first thing we need to do is find the density of modes. The second thing is to findthe mean oocupancy or mean energy of each mode using the Boltzman distribution. What are modes? Youget resonances with the length of the cavity or of the string is equal to an integer of its wavelength
L1 =λ2
L2 = λ L3 =3λ2
or a general expression is given by
λ = n2L, n = 1,2,3, ...
where
ω = 2π f =2πcλ
= nπcL
to do this calculation we need to regard each mode as being equivalent to the simpe harmonic oscillator ofenergy levels at that frequency. The energy of the SHO is given by
ε = n~ω n = 1,2,3, ...
wheren is the number of photons in that energy states.
66
4.2 Density of modes
Lets consider a cavity with sidesL, lets have the waves propgate in thez direction. We have the electricfield and the magnetic field that are orthogonal to each other.There are also two different polarization thatare independent. Lets consider (for example)Ex , next we write down the wave equation
∇2Ex =
(
∂2
∂x2 +∂2
∂y2 +∂2
∂z2
)
Ex =1c2
∂2Ex
∂t2
but we know thatEx(x,y,z, t) = Ex0 sin(ωt)sin
(nxπxL
)
sin(nyπy
L
)
sin(nzπz
L
)
writing this in this form already places boundary conditions. Where they are thatEz vanishes on all thewalls. Now we must substituteEx into the wave equation. We get the following
−π2
L2(n2x +n2
y +n2z) = −ω2
c2
this gives us a relationship between all of then, ω, and the cavity. Thus
n2 =L2
π2c2ω2 n =Lπc
ω n≥ 0
we need to remember that this is in three dimensions. We now want to calculate the density of nodes forthis box.
D(ω)dω = number of modes betweenω,ω+dω
we now need to cosider an octant of a sphere, in fact we want to consider a shell that has a thicknessdnand a radius that we will calln, this shell contains a certain number of nodes. We need to find
D(ω)dω = 218×4πn2dn= πn2dn
the 2 comes about from the two independent modes of polirization, then we need to consider that thesurface area is 1/8 times the surface of the sphere and thedn, thus we have
D(ω) = πn2 dndω
but we know what this isdndω
=Lπc
thus
D(ω) = πL2
π2c2ω2 Lπc
= π(
Lπc
)3
ω2
and we find
D(ω) =V
π2c3ω2
this is a very important result. This is step one of our calculation, we must now go to step two.
67
4.3 Rayleigh-Jeans Theory (1900) classical
By the Equipartition theorem, there is an average energykBT per mode because of the two quadratic termin E2 andB2, this immideatley gives us a result for the energy in the black body spectrum
u(ω) = D(ω)kBT =V
π2c3ω2kBT
if we integrate over the entire spectrum to get the energy we will get the following
uT(ω) =Z ∞
0D(ω)kBTdω =
Vπ2c3kBT
Z ∞
0ω2dω
we can see that this integral diverges. This cannot be true. It was know that the spectrum does not go toinfinity. The classical theory worked well at low frequencies, but not at high frequencies. Then came MaxPlanck who introduced the high frequency limit
uT(ω) ∼ ω2e−~ω/kBT
we now need to show where this came from.
4.4 Planck Theory (1903) Quantum Theory
This results will be the same as the QSHO. Lets consider a modein our box with frequencyω, we wantto know what the average energy at a temperatureT. Lets consider a resevior with one mode in a system(box) and the resevoir is all the other modes. We know that we can solve this using the partition function
Z =∞
∑n=0
e−nε/τ
wheren is the number of photons in the mode and the energy is quantized. Thusε = n~ω, this is really thedifference between the classical theory and quantum theory. Lets find the mean occupancy of that mode
〈n〉 =∑∞
n=0ne−nε/τ
∑∞n=0e−nε/τ
if we writey = e−ε/τ
thus
〈n〉 =∑nyn
∑yn
there is a simple trick to do this
〈n〉 = yddy
ln∞
∑n=0
yn
these two expressions are equivalent. Now we know that
∞
∑n=0
yn =1
1−y
68
and so we are left with
〈n〉 = −yddy
ln(1−y)
thus
〈n〉 =y
1−y=
11/y−1
and thus we get the result
〈n〉 =1
e~ω/τ −1
this is known as thePlanck distribution . There are two points to make, This function takes care of theultraviolet catastrophe
1. Occupancy tends to zero asω goes to zero
2. This whole theory dependend on the fact that we quantized the energy. Thus EM radiation can berepresented by photons of discrete energy. The results implies that the energy is in fact quantized inunits of~ω (“photons”). This was the introduction in Quantum theory.
We can now write the energy density as
u(ω)dω = 〈n〉~ωD(ω)dω
where~ω is th energy of the mode and〈n〉 is the average number of photons with some energy. Thus
u(ω)dω =V~ω3dω
π2c3(e~ω/τ −1)
we can write this more suggestively as
x =~ωτ
this gives
u(ω)dω =V~
π2c3
( τ~
)4(
x2dxex−1
)
thus we can see that the energy goes to the fourth power of the temperature. It will also give us a very usefulscaling argument. The derivative of this function gives us the turning point of the energy distribution. Wefind the maximum to be given by
ddx
(
x2
ex−1)
)
xmax=~ωmax
τ= 2.82
this gives us the following scaling law This results gives uswhat is known as Wien’s law
~ωmax
τ1=
~ωmax
τ2
this enables one to measure temperatures from the black bodyspectrum, e.g stars and pyrometers. Letslook atb this results at the limits.
69
• The classical limit~ω/τ ≪ 1, the discrete energy levels is smeared out. We get the following form
u(ω) =V~ω3
π2c3(~ω/τ)=
Vω2τπ2c3
first we can see that~ drops out and also this is precisely the Rayleight-Jeans result.
• The Quantum limit~ω/τ ≫ 1, energy levels are very discrete
u(ω) =V~ω3
π2c3 e−~ω/τ
this also comes from Wien’s law (1896).
The final thing we need to do is integrate this to find the total energy in the cavity. Thus the total energyin the cavity is given by
uτ =Z ∞
0u(ω)dω
it is convinient to take the formed introduced previoulsy
uτ =V~
π2c3
( τ~
)4 Z ∞
0
x3
ex−1dx
=π2Vτ4
15~3c3 ∝ τ4
where we have saidZ ∞
0
x3
ex−1dx=
π4
15
thus we find the total energy of the cavity is given by
uτ =π2Vτ4
15~3c3
this result is independent of the nature of the cavity or of materials placed inside of it. One othet thing thatwe can work out is the entropy.
4.5 Entropy
We know thatτdσ = du+ pdV
thus, at constant volume we have
dσ =duτ
τwhere
uτ =π2Vτ4
15~3c3 duτ =4π2Vτ3
15~3c3 dτ
and so we find
dσ =4π2Vτ2
15~3c3 dτ
if we now integrate both sides we find
σ =445
π2Vτ3
~3c3
70
4.6 Absorption and Emission of Blackbody radiation
The Energy emmited by a blackbody, where the number of photons emmtitted per unit area per second
N(ω) =14
n(ω)c
we can compare this with the kinetic theory of gases. With this in mind it is simle to write the poweremmited for the blackbody. Where the power emmited per unit area in the frequency range ofω +δω isgiven by
Pl(ω)dω =14
n(ω)c(~ω)dω
now we know that
n(ω)~ω =uT(ω)
Venergy per unit volume
thus the total energy is given by the following
total emmited power=Z ω
0Pl (ω)dω =
14
cZ ω
0
uT(ω)dωV
=14
utot
V
where
utot =π2Vτ4
15~3c3
thus we find
Ptot =π2τ4
60~3c2 = σSBT4
this expressions defines the Stephan-Boltzman constant
σB =π2k4
B
60~3c2 = 5.67×10−8 W
m2K4
this quantity is know very precisely. We will now spend a few minutes talking about absorption andemmision of blackbodies.
4.7 Absorption and Emission
These are the kinds of objects that we might expect to find.
1. Blackbody-is an object that absorbs/radiates all the radiation incident on it. We define this propertyin terms of an absorption coefficient.
a(ω) = 1 aborptivity
we also have what is called a grey bodya(ω) < 1
we also have colored blackbodies
a(ω) depends on frequency
and lastly we have what is called a mirror
a(ω) = 0
given these different properties, how is it that all this objects are at the same temperature in the box?
71
4.7.1 Kirchoff’s Laws
We can define emmisivitye(ω), which is simply the fraction of electromagnetic radiationcompared tothat emitted by a black body emmited at the same temperature.Kirchoff’s law gives
a(ω) = e(ω)
what happens at the microscopic level? If we suppose that we have a solid with many atoms that also havevibration. There are two energy levels. This means that we absorb and emit strongly at some frequencyω.This is know as theprinciple of detailed balance. There are lots of good example of this, i.e a silver teapot should be polished to keep your tea warm for a long time. This is because if the tea pot is very shinythan it will be a poor observer, thus it is also a bad emmiter. Another example is the green house effect.If you have a green house and it is made of glass and inside you have some bushes. What happens is thatUV and optical can pass relatively easy, thus the plants absorb and then re-emit at different frequency.This re-emmited radiation is not able to escape thru the glass because of the glass. Thus the UV is trappedinside the green house. The glass absorbs and emmits strongly in the infrared, hence the green housewarms up. There is also “Global Warming”, this is due to the gases and clouds in the atmosphere. Onclear nights it gets colder than if it is cloudy. The clouds act as an absorbing and emmiting medium forinfrared radiation.
4.8 Phonos: Debye Model
If we think of latice waves then we are really thinking about sound waves, thus they propogate at thevelocity of soundvs, there are three possible independent modes. The first is a longitudinal wave, backand forth motion (spings). An alternative is a transverse vibration, of which there are two (up and down)or perpendicular (in and out). We need to apply the theory of EM radiation to this situation. We need tofigure out the density of modes for this picture.
4.8.1 Density of Modes
There is a distinction between solids and EM radiation. For EM waves, there is an unlimited numer ofpossible modes. On the other hand, sound waves in a solid, thenumber of modes is bounded. Eachatom has three degrees of freedom, thus the maximum number ofmodes for each atom there are 3 modesand forN atoms there are 3N modes. If we consider a single transverse wave, it takes discrete points toidentify that wave. If we try to draw the frequency with only 2points, then this has no meaning. There isa minimum wavelength. We can recall that for photons (2 modes)
D(ω) =Vω2
π2c3
but for phonons (3 modes)
D(ω) =3Vω2
2π2v3s
there is an additional twist. This is cut off atωD, which is the Debye frequency (highest frequency wherewe can have vibrations). Thus
3N =Z ωD
0D(ω)dω =
3V2π2v3
s
Z ωD
0ω2dω =
Vω3D
2π2v3s
72
thus the Debye frequency is given by
ωD =
[
6π2v3s
(
NV
)]1/3
this results is independent of the size of the solid. The velocity of sound is
v3s =
Vω3D
6π2N
if we plug this into the previous equation we find
D(ω) =3Vω2
2π2
6π2N
Vω3D
= 9Nω2
ω3D
thus this scales as∝ Nω2 whereω2 as for photons. The second step is to figure out the populationof thesemodes.
4.9 Total Energy
The total energy at temperatureT is given by
utot(T) =Z ωD
0D(ω)~ω〈n〉dω
where〈n〉 is the average number of phonos in a mode,~ω is the phonon energy, andD(ω) is the densityof modes. Thus we find
utot(T) =9N~
ω3D
Z ωD
0
ω3dωe~ω/kBT −1
so we can write this integral in a dimensionless form
x =~ωkBT
, xD =~ωD
kBT=
kBTD
kBT=
θD
T
whereθD is the “Debye temperature”
θD =~ωD
kBω =
xkBT~
dω =kBT~
dx
thus we will get the following
u(T) =9N(kBT)4
~3ω3D
Z xD
0
x3
ex−1dx
whereZ xD
0
x3
ex−1dx
is dimension-less, this can only be solved numerically or bylookin at the extremes (limits).
73
4.9.1 Low temperature
xD ≫ 1 orkBT ≪ ~ωD = kBθD , we can replacexD by infinity in the upper limit, and hence
Z ∞
0
x3
ex−1dx=
π4
15
thus in this limit
u(T) =9N(kBT)4
~3ω3D
π4
15
thus
u(T) =3π4N(kBT)4
5(~ωD)3
this is the total energy of a solid at low temperatures. It is more useful to talk about the heat capacity
CV =
(
∂U∂T
)
V=
12π4NkB
5
(
kBT~ωD
)3
thus
CV =12Nπ4
5kB
(
TθD
)3
this is called the DebyeT3 law. All non-magnetic insulator obey this law. But metals donot obey thisbecause they have electronic contribution.
4.9.2 High Temperature
xD ≪ 1 orkBT ≫ ~ωD or T ≫ θD , this is a simple case
Z xD
0
x3
ex−1dx
thus we can expandex as 1+xZ xD
0x2dx=
x3D
3and we find the total energy to be
u(T) =9N(kBT)4
~3ω3D
x3D
3
which yieldsu(T) = 3NkBT
wher 3N is the total number of atoms in the crystal. We haveN atoms that act as 1D harminic oscillator,this is in fact a 3D problem. This is the equipartition theorem applied toN 3 dimensional simple harmonicoscilator. We can easily derive the heat capacity
CV = 3NkB
this is a very important results, this is known in chemistry as theDulong and Petit lawwhich is
≈ 25 joule deg−1 mole−1
74
we can also plotCV vs the temperature we find, at low temperature we have the famousT3 law and at hightemperature we find that this is independent of temperature.θD is a measured quantity, a few examplesare given as
Indium 108K
Copper 343KDiamond 1500K
there is a clear temperature dependence and hardness of the material. Hard materials are “stiff” thereforethey have a high value of spring constant, thus highωD ⇒ highθD. We should expect that there shouldalso be a contribution from the electric contribution to theheat capacity. The metals have free electronsthat apperently do not contribute significantly toCV at room temperature.
4.10 Fluctuations in Phonon Energy for Temperatures≪ θD in T3
Limit
We have a resevoir and a system, we want to see how the fluctuations scale with temperature. We rememberthat
〈(∆u)2〉 = −(
∂u∂β
)
= τ2(
∂u∂τ
)
= (kBT)2(
∂ukB∂T
)
= kBT2CV
we need to recall the expressio for the energy in this limit
u(T) =3π4(kBT)4N
5(~ωD)3 =3π4
5kBT4N
θ3D
we know that the energy is in the form ofu = AT4
thus the heat capacity is
CV =
(
∂u∂T
)
= 4AT3
and now we can write
〈(∆u)2〉u2 =
kBT24AT3
uAT4 =4kBT
u=
20kBTθ3D
3π4kBT4N=
203π4
1N
θ3D
T3 =
(
203π4
)(
θD
T
)3 1N
if we take a small thin film withl = 100µm w = 10µm, andh = 0.1µm which gives us a volume of
V = 10−16 m3
where the typical density is≈ 1028 m3, thus we hav e approximatelyN ≈ 1012 atoms, and also
θD ≈ 300 K T = 0.01 K
if we put all these number in we find
〈(∆u)2〉u2 ≈ 1
15
(
3000.01
)3
×10−12 ≈ 2
thus we have large fluctuations in the energy, in fact, fluctuations results break down because the energyof the system cannot be negative.
75
4.11 Summary
We started talking about electromagnetic radiation. The first thing was the density of modes
D(ω) =Vω2
π2C3
we then found the famous Planck distribution
〈n〉 =1
e~ω/kBT −1
and the energy density is given by
u(ω) = 〈n〉~ωD(ω) =V~ω3
π2c3(~ω/τ)
and the total energy is given by
uτ =π2Vτ4
15~3c3
We also found the total power to be given as
P = σSBT4
where
σSB=π2k4
B
60~3c2 = 5.678×10−8 Wm−2K−4
we also derived Kirchoffs lawsa(ω) = e(ω)
we also discussed phonos
CV =12π4
5NkB
(
TθD
)3
T ≪ θD
andCV = 3NkB T ≫ θD
4.12 Problems and Solutions
Problem # 1A wire of a radiusr0 is coincident with the axis of a metal cylinder of radiusR and lengthL. The
wire is maintained at a positive potentialV with respect to the cylinder. The whole system is at some hightemperatureT. As a result, electrons emitted from the hot metals form a dilute gas filling the cylindricalcontainer and in equilibrium with it. The density of these electrons is so low that their mutual electrostaticinteractions can be neglected.
76
a) Use Gauss’s theorem to obtain an expression for the electrostatic field which exists at points at a radialdistancer from the wire (r0 < r < R). The cylinder of lengthL may be assumed to be very long sothat end effects are negligable.
We know that Gauss’s law is given byZ
E ·da=qenc
ε0
where we know thatZ
da= A = 2πrL
thus we find the electric field to be given by
E =qenc
2πrLε0
but we know that the charge enclosed can be found through the potential
V =
Z
E ·dl =qenc
2πLε0
Z R
r0
1r
dr =qenc
2πLε0ln
(
Rr0
)
thus
qenc=2πLε0
ln(R/r0)V
thus we find the electric field to be given by
E =V
r ln(R/r0)(4.1)
b) In thermal equilibrium, the electrons form a gas of variabledensity which fills the entire space betweenthe wire and cylinder. Using the result of part (a), find the dependence of the electric charge density(relative density) on the radial distancer.
We know that the relative density is given by the Boltzman factor, since we can assume that there areN1
andN2 particles having energyε1 andε2 respectively then we can write the relative density as
ρ =ρ(r1)
ρ(r2)= e−(ε(r1)−ε(r2))/τ
77
we have proved this result elsewhere. Thus we need to find whatthe energy is at distancer1 andr2. Weknow that the energy due to some charge and potential is givenby
ε(r) =p2
2m−qV(r)
thus we find that∆ε = q[V(r2)−V(r1)]
if we let
V(r) = CZ r
r0
1r
dr = C ln
(
rr0
)
thus
V(r1) = C ln
(
r1
r0
)
V(r2) = C ln
(
r2
r0
)
this allows us to find thet the change in energy is given by
∆ε = Cq
[
ln
(
r2
r0
)
− ln
(
r1
r0
)]
where we know thatC is given from Equation 1 as
C =V
ln(R/r0)
thus we find the relative density to be given as
ρ =ρ(r1)
ρ(r2)= e−∆ε/τ = e
−Cq[
ln(
r2r0
)
−ln(
r1r0
)]
/τ
if we choose appropriate radiusr1 = r0 andr2 = r2 we find the the relative density is given by
ρ =ρ(r0)
ρ(r2)= e
− Vln(R/r0)
q[
ln(
r2r0
)]
/τ
Problem # 2 Zipper ProblemA zipper hasN links; each link has a state in which it is closed with energy 0and a state in which it
is open with energyε. We require, however, that the zipper can only unzip from the left end, and that thelink numbers can only open if all links to the left (1,2,...,s-1) are already open.
a) Show that the partition function can be summed in the form
Z =1−exp(−(N+1)ε/τ]
1−exp(−ε/τ)
First of all, we know that the number of states is given bys= (1,2,3, ...,s−1) and we also know that thetotal energy is given byεs = nε, thus we can write the partition function as
Z =N
∑s=0
e−εs/τ =N
∑n=0
e−nε/τ =N
∑n=0
(e−ε/τ)n
78
this resembles a geometric series of the form
k
∑n=0
arn = a1− rk+1
1− r
thus we can see that the partition function is simply
Z =1−exp(−(N+1)ε/τ]
1−exp(−ε/τ)
b) In the limit ε ≫ τ, find the average number of open links. The model is a very simplified model of theunwinding of two-stranded DNA molecules.
We know that at this limit, the partition function reduces to
Z =1
1−exp(−ε/τ)
and we know that the average number of open links is given by
〈N〉 =N
∑n=0
np(l) =∑N
n=0ne−nε/τ
Z
if we expand the numerator to the first few terms we find
〈N〉 = (e−ε/τ +2e−2ε/τ +3e−3ε/τ)(1−e−ε/τ)
if we letx = e−ε/τ
we find that the average number of open links is given by
〈N〉 = (x+2x2 +3x3)(1−x)
Problem # 3 Elasticity of PolymersThe hermodynamic identity for a one-dimensional system is
τdσ = dU− f dl
when f is the external force exherted on the line anddl is the extension of the line. By analogy with
p = τ(
∂σ∂V
)
U
we form
f = −τ(
∂σ∂l
)
U
the direction of the force is opposite to the conventional ofthe pressure, this is due to the fact that(∂σ/∂l )is always negative.
We consider a polymeric chain ofN links each of lengthρ, with each link equally likely to be directedto the right and to the left.
79
a) Show that the number of arrangements that give a head-to-tail length of l = 2|s|ρ is
g(N,−s)+g(N,s) =2N!
(12N+s)!(1
2N−s)!
we know that the multiplicity is defined as
g(N,s) =N!
(12N+s)!(1
2N−s)!
we can also see thatl = Nρ = 2|s|ρ 2|s| = N
we can also see that this system can only be in two possible states
g(N,−s) state 1 g(N,s) state 2
we can add these two multiplicity functions (which gives thetotal multiplicity of the system), we simplyget
g(N,−s)+g(N,s) =2N!
(12N+s)!(1
2N−s)!
b) For |s| ≪ N show that
σ(l) = ln[2g(N,0)]− l2
2Nρ2
since we know that the entropy is defined as
σ = ln[g(N,s)]
and the Sterling approximation to the multiplicity is givenby
g(N,s) ≈ g(N,0)e−2s2/N
and since we know that
s=l
2ρwe find that the multiplicity is given as
g(N,s) = g(N,0)e− l2
2Nρ2
thus the entropy yields
σ = ln[g(N,0]− 2s2
Nin our case we know that
g(N,±s) = g(N,−s)+g(N,s) = 2g(N,s)
thus we find that the entropy for this system is given by
σ = ln[2g(N,0)]− l2
2Nρ2
80
c) Show that the force at extensionl is
f =lτ
Nρ2
The force is proportional to the temperature. The force arises because the polymer wants to curl up: theentropy is higher in a random coil than in an uncoiled configuration. Warming a rubber band makes itcontract; warming a steel wire makes it expand.
Since we know what the entropy is for our system and we also know that the force is given by
f = −τ(
∂σ∂l
)
we simply need to differentiate the entropy as a functio of length, which yields
∂σ∂l
= − lNρ2
Substituting this into the force yields
f =lτ
Nρ2
Problem # 4 Earth’s Steady State TemperatureThe surface temperature of the Sun isT0 = 5500 K; its radius isRsun= 7×1010 cm while the radius of
the Earth isRearth= 6.37×108 cm. The mean distance between the Sun and the Earth isr = 1.5×1013 cm.In the first approximation one can assume that both the Sun andthe Earth absorb all electromagneticradiation incident upon them. The Earth has reached a steadystate so that its mean temperatureT doesnot change in time in spite of the fact that the Earth constantly absorbs and emits radiation.
a) Find an approximate expression for the temperatureT of the Earth in terms of the astronomicalparamters mentioned above.
We know that the power recieved at the Earth from the sun is given by
P(earth) = FsunAearth =Lsun
4πr2πR2earth =
Lsun
4r2 R2earth
wherer is the distance from the Sun to the Earthr = 1 AU. We also know that the luminosity of an objectis given by
Lsun= 4πR2sunσT4
sun
thus the power the Earth receives from the sun is given by
P(earth) =πR2
sunσT4sunR
2earth
r2
we know that the power emmited by the Earth is given
P(earth) = 4πR2earthσT4
earth
setting these two expressions together we find that the temperature of the Earth is given by
πR2sunσT4
sunR2earth
r2 = 4πR2earthσT4
earth
81
thus
Tearth = Tsun
√
Rsun
2r
b) Calculate this temperatureT numerically.
Since we know thatRsun= 7×108 m andr = 1.5×1011 m andTsun= 5500 K we find the temperature ofthe Earth to be
Tearth = 5500 K
√
(7×108)
2(1.5×1011)
this is simplyTearth = 265.67 K
Problem # 5 Number of Thermal PhotonsShow that the number of photons∑〈sn〉 in equilibrium at temperatureτ in a cavity of volumeV is
N = 2.404π−2V(τ/~c)3
where we know that the entropy is given by
σ =
(
4π2V45
)
( τ~c
)3
whenceσ/N ≃ 3.602. It is believed that the total number of photons in the universe is 108 larger than thetotal number of nucleons (protons, neutrons). Because bothentropies are of the order of the respectivenumber of particles, from
σ = N
[
ln(nQ/n)+52
]
the photons provide the dominent contribution to the entropy of the universe, although the particles domi-nate the total energy. We believe that the entropy of the photons is essentially constant, so that the entropyof the universe is approximately constant with time.
We know that
〈sn〉 =1
e~ωn/τ −1
thus we can see that
N = ∑n〈sn〉 = ∑
n
1
e~ωn/τ −1
where the sum is over the triplet of integersnx,ny,nz. Positive integers alone will describe all independentmodes. We replace the sum overnx,ny,nz by an integral over the volume elementdnx,dny,dnz in thespace of the mode indices. That is we set
N =Z ∞
0
n2dn
e~ωn/τ −1dΩ
82
wheredΩ is the solid angle given by
dΩ =Z 2π
0dφ
Z π
0sinθdθ = 4π
thus we find
N = 4πZ ∞
0
n2dn
e~ωn/τ −1to make our lives easier, lets just integrate over one octant, this gives us
N =4π8
Z ∞
0
n2dn
e~ωn/τ −1
since we know that
ωn =nπcL
n2 =
(
Lωn
πc
)2
dn=Lπc
dω
we are now able to write
N =π2
Z ∞
0
n2dn
enπc~/Lτ −1if we introduce the unitless quantity
x =π~cnLτ
n =Lτ
π~cx dn=
Lτ~πc
dx
using these identities we are now able to write
N =
(
Lτ~c
)3 1π2
Z ∞
0
x2
ex−1dx= π−2V
( τ~c
)3Z ∞
0
x2
ex−1dx
where we have looked up the solution toZ ∞
0
x2
ex−1dx≈ 2.404
thus we find
N = 2.404π−2V( τ
~c
)3
Problem # 6 Average Temperature of the Interior of the Sun
a) Estimate by a dimensional argument or otherwise the order ofmagnitude of the gravitational self-energy of the Sun, withMsun = 2×1033 g andRsun = 7×1010 cm. The gravitational constantGis 6.6× 10−8 dyne cm2g−2. The self-energy will be negative referred to atoms at rest at infiniteseperation.
The energy due to a gravitational field produced my some massM is given by
U = −GM2
r
given the quantities forM,G andRwe find
U = −GM2
r≈ −3.81×1041 Joules
83
b) Assume that the total thermal kinetic energy of the atoms in the Sun is equal to−12 times the gravita-
tional energy
K = −12U Virial Theorem
this is a result of the Virial Theorem in mechanics. Estimatethe average temperature of the Sun.Take the number of particles as 1× 1057. This estimate gives somewhat too low a temperature,because the density of the Sun is far from uniform. “The rangein central temperature for differentstars, excluding only those composed of degenarate matter for which the law of perfect gases doesnot hold (white dwarfs) and those which have excessively small average densities (giants and supergiants), is between 1.5 and 3.0×107 degrees.”
Since we know that the total therma kinetic energy is given by
K = NkBT = −12U = 1.9×1041 joules
thus we find that the temperature will be given as
T = − U2NkB
=1.9×1041 joules
NkB
since we know the number of particles and what the constantkB we find that the temperature is
T =1.9×1041 joules
1×1057 ·1.38×10−23 joulesK ≈ 7.24×106 K
Problem # 7 Angular Distribution of Radiant Energy Flux
a) Show that the spectral density of the radiant energy flux thatarrives in the solid anledΩ is cuω cosθ ·dΩ/4π, whereθ is the angle the normal to the unit area makes with the incident ray, anduω is theenergy denisty per unit frequency range.
84
Since we know that the differential energydE is given by
dE = cdtuωdΩdA
where we also know that the incident beam will not contributeanything to the flux if the angle ofincidence is 90 degrees orπ/2. Thus I know that my differential areadA is really given bydA= cosθdA,this will account for any angle dependence on the incidence of the beam. Thus we find
dE = cdtuωdΩcosθdA
or this can also be written asdEdt
= cuω cosθdΩdA
we must also normalize this over all solid angle 4π ster-radians, thus we find
dEdt
= cuω cosθdAdΩ4π
we also know that the flux is defined as
F(ω) =dE
dtdAthus we find that the flux density is given as
F(ω) = cuω cosθdΩ4π
85
b) Show that the sum of this quantity over all incident rays is (1/4)cuω.
We know that the sum of this quantity will be given by the integral
Ftot =cuω4π
Z 2π
0dφ
Z π/2
0sinθcosθdθ
this is given as
Ftot =cuω2
Z π/2
0sinθcosθdθ
we know that the integralZ π/2
0sinθcosθdθ =
12
thus we find the total flux integrated over the solid angle is given by
F =14
cuω
Problem # 8 Free energy of a photon gas
a) Show that the partition function of a photon gas is given by
Z = ∏n
[1−e−~ωn/τ]−1
where the product is over the modesn.
We know that the partition function for a photon in a moden is given by
Z =∞
∑n
e−~ωn/τ
This sum is of the form∑xn with x≡ exp(−~πc/τL). Becausex is smaller than 1, the infinite series maybe summed and has the value
Z =1
1−e−~ωn/τ = [1−e−~ωn/τ]−1
since we know that for many photons in a gas the partition function can be written as
Zn = Z1Z2Z3...Zn
since this aredistinguishablephotons in a given energy staten. Thus the partition can be expressed as aproduct of the form
Z = ∏n
[1−e−~ωn/τ]−1
b) The Helmholtz free energy is found directly as
F = τ∑n
ln[1−e−~ωn/τ]
86
Transform the sum to an integral; integrate by parts to find
F = − π2Vτ4
45~3c3
turning this sum into an integral yields
F = 2× 18
τ4πZ ∞
0n2 ln[1−e−~ωn/τ]dn
where the 2 comes from the two polirizations, the 1/8 comes from the fact that we are only integrat-ing over one octant and the 4π comes from integrating over the entire solid angle of a sphere. Wealso know that
ωn =πncL
thus the free energy (in integral form) is
F = τπZ ∞
0n2[1−e−~nπc/Lτ]dn
letting
x =~nπcLτ
dn=Lτ
~nπcdx n2 =
(
Lτ~πc
)2
x2
yields
F =τ
π2
(
Lτ~c
)3 Z ∞
0x2 ln[1−e−x]dx
integration by parts yields
F =τ
π2
(
Lτ~c
)3[[x3
31−e−x
]∞
0− 1
3
Z ∞
0
x3
1−e−xdx
]
\right]since we know that
x3
3ln[1−e−x]∞0 → 0
Z ∞
0
x3
1−e−xdx=π4
15
we find that the free energy is
F = − π2Vτ4
45~3c3
Problem # 9 Heat capacity of liquid4He at low temperatures
The velocity of longitudinal sound waves in liquid4He at temperatures below 0.6 K is2.383×104 cm s−1. There are no transverse sound waves in the liquid. The density is 0.145 gm cm−3.
a) Calculate the Debye temperature.
87
In order to calculate the Debye temperature for4He we must begin with the total number of elastic modesequalingN
4π8
Z nmax
0n2dn= N
we find thatπ6
n3D = N nD =
(
6Nπ
)1/3
the thermal energy of the phonon is
U = ∑〈εn〉 = ∑〈sn〉~ωn = ∑ ~ωn
e~ωn/τ −1
for this problem we find the thermal energy to be given as
U =4π8
Z nD
0
~ωn
e~ωn/τ −1n2dn
if we letωD =
πvnD
Lwe find the energy to be given as
U =~πvnD
Lπ2
Z nD
0
n2D
e~πnDv/Lτ −1dn
if we let
xD =π~nDv
Lτ=
π~vLτ
(
6Nπ
)1/3
=θT
=kBθ
τwhereθ is called theDebye temperatureexpressed as
θ =~vkB
(
6π2NV
)1/3
since we know that
N =Mtot
m1
NV
=Mtot
V1
m1=
ρm1
wherem1 is the atomic mass unit of a helium atom. Since we know
~ = 1.05×10−34Js
kB = 1.38×10−23J/K
m1 = 4×1.66×10−24g
v = 2.383×104 cm s−1
ρ = 0.145 gm cm−3
we find the Debye temperature to be given as
θ = 19.71 K
88
b) Calculate the heat capacity per gram on the Debye theory and compare with the experimental valueCV = 0.0204×T3, in J g−1K−1. TheT3 dependence of the experimental value suggest that phonosare the most important excitations in liquid4He below 0.6 K. Note that the experimental value hasbeen expressed per gram of liquid.
The energy in the low temperature limit for this case is givenas
U =π4Nτ4
5(kBθ)3
we can see that the heat capacity is given as
CV =
(
∂U∂τ
)
V=
4π4Nτ3
5(kBθ)3 =45
π4kBNT3
θ3
to find the heat capacity per gram we use
CV
g=
4π4kB
5θ3
Nmass
×T3
whereN
mass=
1m1
wherem1 is the atomic mass unit for helium, thus we find
CV
g=
4π4kB
5θ3m1×T3 = 0.0211 J g−1K−3
the error is given by∆CV ≈ 3.3%
89
Chapter 5
Chemical Potential and the Gibbs Distribution
This happens in a system that can change both energy and number of particles.
Road map
1. Definition of chemical potential
2. Thermodynamic identity (expanded) and the free energy
3. Tow kind of chemical potential: Internal and external chemical potential
4. Gibbs factor and the grand partition function.
5.1 Definition of Chemical Potential
Lets consider two system in thermal and diffusive contact. We have two systems that are seperated by apermeable material where particles can go through it. We have
N1,u1 and N2,u2
whereu = u1+u2 N = N1+N2
these are both constant. We will proceed by writing down the most probable configuration MPC occurswhen
g1(N1,u1)g2(N2,u2)
is a maximum. If we diferentiate this, we know that
d(g1,g2) =
[(
∂g1
∂N1
)
dN1+
(
∂g1
∂u1
)
du1
]
g2+
[(
∂g2
∂N2
)
dN2+
(
∂g2
∂u2
)
du2
]
g1
we will also now saydu1 = −du2 dN1 = −dN2
if we now divided byg1,g2 we find[
1g1
(
∂g1
∂N1
)
− 1g2
(
∂g2
∂N2
)]
dN1 +
[
1g2
(
∂g1
∂u1
)
− 1g2
(
∂g2
∂u2
)]
du1 = 0
90
but now we note the following
1g1
(
∂g1
∂u1
)
=
(
∂ lng1
∂N1
)
=
(
∂σ1
∂N1
)
if we now this for all these terms we get the following
1g2
(
∂g2
∂u2
)
=
(
∂ lng2
∂N2
)
=
(
∂σ2
∂N2
)
thus we find[(
∂σ1
∂N1
)
−(
∂σ2
∂N2
)]
dN1 +
[(
∂σ1
∂u1
)
−(
∂σ2
∂u2
)]
du1 = 0
now we know thatdN1 anddu1 are independent. Hence each term in the brackets is independent of 0.(
∂σ1
∂u1
)
=
(
∂σ2
∂u2
)
⇒ 1τ1
=1τ2
but we find a new term(
∂σ1
∂N1
)
=
(
∂σ2
∂N2
)
thus we can now define the chemical potential as
−µτ
=
(
∂σ∂N
)
u
this is thechemical potential. The chemical potential has dimensions of energy. We know that particleswill diffuse from a higherµ until µ1 = µ2 at thermal equilibrium. We can removeδN particles from 2 andadd to 1. Thus
δN1 = −δN2 = δN > 0
and
δσ = δσ1+δσ2 =
(
∂σ1
∂N1
)
δN−(
∂σ2
∂N2
)
∂N =(
−µ1
τ+
µ2
τ
)
δN
if initiallyµ2 > µ1
thus the entropy would increase andδσ is positive and thus we are moving to thermal equilibrium. Asaconsequence particles will flow from higherµ to lowerµ until they reach equilibrium.
5.2 Thermodynamic Identity and Definition of µ
We know that we can write the entropy as
σ = σ(u,V,N)
if we take the derivative
dσ =
(
∂σ∂u
)
du+
(
∂σ∂V
)
dV+
(
∂σ∂N
)
dN
91
thus
dσ =1τ
du+pτ
dV− µτ
dN
we ca now simply write thisdu= τdσ− pdV+µdN
this is now ourthermodynamic identity.We have just derived the thermodynamic identity. This is really the first law of thermodynamics, it
also implies conservation of energy. This also gives us a definition of the chemical potential
µ=
(
∂u∂N
)
σ,V
lets recall the Helmholtz free energyF = u− τσ
if we differentiate we find
dF = du− τdσ−σdτ= τdσ− pdV+µdN− τdσ−σdτ
dF = −σdτ− pdV+µdN
thus we finddF = −σdτ− pdV+µdN
this makes the free energy a function of the entropy, temperature, and chemical potential. Hence
µ=
(
∂F∂N
)
τ,V
what we see is that the two different definitions of the chemical potential allow you to considerµ as theincrease in the internal energyu when 1 particle is added to the system at constantσ andV. It is also theincrease in the Helmholts free energy to the system at constant τ andV. You can have both internal andexternal contributions to the chemical potentials.
5.3 Internal and Total Chemical Potential.
This is two say that there are two kind (contribution) of chemical potential,namely, internal and external.Lets suppose that we have two boxes of gas that have initiallyµ1,N1,τ1 and the second hasµ2,N2,τ2,initially
∆µ(initial ) = µ2(initial )−µ1(initial ) > 0
thus we have a flow of particles from 2 to 1. Lets also assume that they have a chargeq. Lets now connecta battery between the two boxes with an EMF∆V. We can assume that the chemical potential is fixed. Ifwe add a potential to them the chemical potential will increase. If we consider a few of the energy levelsof the system. If we now add a potential than we simply raise all of the energy levels, so that subsequentlywe have a situation where we have simply increased the energyof all the particles by an amountq∆V.We thus have both an internal and external contribution to the chemical potential. We apply a voltagedifference∆V to cancel∆µ, thus
q∆V = ∆µ(initial )
92
we can see thatµ1( f inal) = µ1(initial )+∆µ(initial )= µ2(initial ) = µ2( f inal)
the chemical potential is simply a potential energy. This reenforces the notion that the chemical potentialis nothing more than a potential energy. So the total chemical potential is
µtot = µint +µext
whereµext is the potential energy of the particle due to an external source of energy andµint is the energyper particle without the external energy. The externla contribution could be, electric, magnetic, gravitati-nal, etc. The total potential is often referred to as the electric-chemical potential if the external contributioncomes from electric potential.
5.4 The Lead-Acid Battery
We must understand that the flow of charge is carried by the ions in the solution (chemical potential). Thehydrogen ions flow to the right and the sulfate ions flow to the left. The chemical reactions are as follows
Pb+SO4 → PbSO4+2e− cathode
where the electrons flow thru the external circuits. At the anode we have the following
PbO2+2H+ +H2SO4+2e− → PbS04+2H2O anode
the net result is that we use up the lead sulfate and produce water. The electrons come from the externalcircuit. The cathode is a sink for sulfate ions, therefore the chemical potential is lower at the cathode (leadsurface)µ(SO−−
4 ) < µ(SO−−4 ) in the electrolyte. If we sketch the chemical potential, we can see that there
is a step given by∆µ(SO−−4 ). We dont know what this is, this is something we must measure.The same
thing happens at the anode, because the anode is now a sink forthe hydrogen atoms. So that the chemicalpotentialµ(H+) at Pb surface is less thanµ(H+) in electrolyte. Thus there is a step at∆µ(H+). Thechemical potential steps drive the ions toward the electrodes and hence the current through the resistor,or more generally, your external circuit. What happens if wedisconnect the external circuit? The currentstops, obviously, how can we stop the chemical reactions? What will happen is that electrons will leave theanode into the electrolyte until they build up a potential difference that exactly cancels chemical potentialand make the total chemical potential zero, this happens when
∆µ(H+)+e∆V+ = 0
Electrons enter the cathode until the total chemical potential is zero
∆µ(SO−−4 )+2e∆V− = 0
when it is all finished we will have∆V = ∆V+ +∆V−
which is our total potential difference between the two terminals. Where the terms on the right are usuallyreferred to as the half-cell potentials, where
∆V− = −0.4V ∆V+ = 1.6V
and∆V = 2.0V
We can see that the battery takes chemical energy to drive electric circuits. The electric chemical potentialis zero to produce zero current flow. The charging process reverses the chemistry and discharge processand it takes sulfate ions and produces sulfuric acid. RemovesSO4 from the electrodes to produce sulfuricacid.
93
5.5 The Gibbs Factor
We know thatτ gives us the Boltzman factor and we will now see that the chemical potential gives us theGibbs factor. We assume that we have a resevior in thermal contact with a system. Where the number ofparticles in the resevoir isN0−N and energyu0− εl . The system is diffusive and in thermal contact. Wewant to find the ensemble average, where the system hasN particles with energyεl ,N. We will start of byassuming that there is only one state that has energyεl so that we have a non-degenarate system. Thus themultiplicity is equal to 1. We can say that the probability that this is in some stateεl with N particles isproportional to multiplicity of the resevoir
P ∝ g(N0−N,u0− εl )×1
i.eP(N,εl) = Ag(N0−N,u0− εl)
we can now do a first order Taylor expansion
lng(N0−N,u0− εl ) = lng(N0,u0)−(
∂ lng∂N
)
u0
N−(
∂ lng∂u
)
N0
εl
thus we can see that
lng(N0−N,u0− εl ) = lng(N0,u0)+µNτ
− εl
τif we now take exponetial of both sides we find
g(N0−N,u0− εl ) = g(N0,u0)e(Nµ−εl )/τ
thus we find the probability is given as
P(N,εl) = Ad(N0,u0)e(Nµ−εl )/τ = Be(Nµ−εl )/τ
this is exactly the form of the Boltzman factor, except that we now have the extra term given by thechemical potential. We can now make a couple of immediate conclusions. You first need the ratios of theprobability
P(N1,ε1)
P(N2,ε2)=
e(N1µ−ε1)/τ
e(N2µ−ε2)/τ
we sometimes get degeneracies, where we can use the same trick as for the Boltzman factor. If we supposethat we have energyρ1 andε1 andρ2 andε2. Thus the ratios of the probabilities of the system havingenergiesε1 andε2 is given by
W(N1,ε1)
W(N2,ε2)=
ρ1e(N1µ−ε1)/τ
ρ2e(N2µ−ε2)/τ
where we now weight the factors.What isB? We need to sum
N0
∑N=0
∑εl
P(N,εl) = BN0
∑N=0
∑εl
e(Nµ−εl )/τ = 1
thus
B =1ζ
where ζ =N0
∑N=0
∑εl
e(Nµ−εl )/τ
94
this we call theGrand Partition Function or the Grand Sum. Finally we have the results for the Gibbsfactor
P(N,εl) =e(Nµ−εl )/τ
ζthis is known as the Gibbs factor it resembles the the Boltzman factor. We want to find the average numberof particles and the average energy of the system.
Mean Number of particles 〈N〉We know that the average nuber of particles is given as
〈N〉 =∑N ∑l Ne(Nµ−εl )/τ
ζ
this is the standard result for finding the average number of particles. This is simple if we make somealgabreac manipulation. Consider
∂ζ∂µ
=Nτ
ζ =1τ
(
∑N
∑l
N(Nµ−εl )/τ
)
=ζ〈N〉
τ
what we see is that the average value ofN is given by
〈N〉 =τζ
∂ζ∂µ
= τ∂ lnζ∂µ
we will also calculate the average energy. The mea energy〈ε〉
〈ε〉 =∑N ∑l εl ∑l Neβ(Nµ−εl )
ζ
lets consider the following derivative
∂ζ∂β
= ∑N
∑l
(Nµ− εl )eβ(Nµ−εl ) = [〈N〉µ−〈ε〉]ζ
thus
〈N〉µ−〈ε〉 =1ζ
∂ζ∂β
=∂ lnζ∂β
and we find the average energy to be given as
〈ε〉 = 〈N〉µ− ∂ lnζ∂β
= µτ∂ lnζ∂µ
− ∂ lnζ∂β
and finally
〈ε〉 = µτ∂ lnζ∂µ
− ∂ lnζ∂β
a result from all this is for the fluctuations
〈(∆N)2〉 = τ∂〈N〉
∂µ
this is left as a homework assigment.
95
5.6 Summary
This are some very important relationships
dσ(u,V,N) =
(
∂σ∂µ
)
V,Ndµ+
(
∂σ∂V
)
µ,NdV+
(
∂σ∂N
)
V,µdN
=duτ
+Pτ
dV− µτ
dN
this gives us the following identitydµ= τdσ−PdV−µdN
and also
dF = d(u− τσ) = du− τdσ−σdτ= −σdτ− pdV+µdN
if we now make a table of all these results
σ(u,V,N) u(σ,V,N) F(τ,V,N)
τ 1τ =
(
∂σ∂µ
)
V,Nτ =
(
∂µ∂σ
)
V,N
p pτ =
(
∂σ∂V
)
µ,N−p =
(
∂µ∂V
)
σ,N−p =
(
∂F∂V
)
τ,N
µ −µτ =
(
∂σ∂N
)
µ,Vµ=
(
∂µ∂N
)
σ,Vµ=
(
∂F∂N
)
τ,V
We then talked about the total chemical potenial
µtot = µint +µext
we then talked about the Gibbs factor
P(N,εl) =e(Nµ−εl )/τ
ζ
whereζ is the grand sumζ = ∑
N∑l
e(Nµ−εl )/τ
and we found the average number of particles
〈N〉 =τζ
∂ζ∂µ
= τ∂ lnζ∂µ
and the average energy is given as
〈ε〉 = µτ∂ lnζ∂µ
− ∂ lnζ∂β
and the fluctuations are given by
〈(∆N)2〉 = τ∂〈N〉
∂µ
96
5.7 Problems and Solutions
Problem # 1 States of positive and negative ionization
Consider a lattice of fixed hydrogen atoms; suppose that eachatom can exist in four states:
state Numbe o f electrons Energy
ground 1 −12∆
Positive ion 0 −12δ
Negative ion 2 12δ
Excited 1 12∆
Find the condition that the average number of electrons per atom be unity. The condition will involveδ,λ, andτ.
The grand partion function is given as
Z = ∑e(Nµ−εn)/τ = e(µ+ 12∆)/τ +e
12δ/τ +e(2µ− 1
2δ)/τ +e(µ− 12∆)/τ
if we letλ = eµ/τ
which is the absolute activity, allows us to simplify the partition function to
Z = e12∆/τλ+e
12δ/τ +e−
12δ/τλ2+e−
12∆/τλ
Z = λ[e12∆/τ +e−
12∆/τ]+e
12δ/τ +e−
12δ/τλ2
we know that the average number of particles is given by
〈N〉 = λ∂
∂λlnZ = λ
∂∂λ
ln[
λ[
e12∆/τ +e−
12∆/τ
]
+e12δ/τ +e−
12δ/τλ2
]
we know that
∂∂λ
ln[
λ[e12∆/τ +e−
12∆/τ]+e
12δ/τ +e−
12δ/τλ2
]
=e
12∆/τ +e−
12∆/τ +2λe−
12δ/τ
λ[e12∆/τ +e−
12∆/τ]+e
12δ/τ +e−
12δ/τλ
thus we find that the average number of particles is
〈N〉 = λ
[
e12∆/τ +e−
12∆/τ +2λe−
12δ/τ
λ[e12∆/τ +e−
12∆/τ]+e
12δ/τ +e−
12δ/τλ2
]
setting this expression equal to 1 yields
λ
[
e12∆/τ +e−
12∆/τ +2λe−
12δ/τ
λ[e12∆/τ +e−
12∆/τ]+e
12δ/τ +e−
12δ/τλ2
]
= 1
thusλ[e
12∆/τ +e−
12∆/τ +2λe−
12δ/τ] = λ[e
12∆/τ +e−
12∆/τ]+e
12δ/τ +e−
12δ/τλ2
97
2λ2e−12δ/τ = e
12δ/τ +λ2e−
12δ/τ
we findλ2e−
12δ/τ = e
12δ/τ
thusλ = eδ/2τ
Problem # 2 Carbon monoxide poisoning
In carbon monoxide poisoning the CO replaces the O2 adsorbed on hemoglobin (Hb) molecules in theblood. To show the effect, consider a model for which each adsorption site on a heme may be vacant ormay be occupied either with energyεA by one molecule O2 or with energyεB by one molecule CO. LetNfixed heme sites be in equilibrium with O2 and CO in the gas phases at concentrations such that theactivities areλ(O2) = 1×10−5 andλ(CO) = 1×10−7, all at body temperature 370 C. Neglect any spinmultiplicity factors.
a) First consider the system in the absence of CO. EvaluateεA such that 90 percent of the Hb sites areoccupied by O2. Express the answer in eV per O2.
We know that the grand partition function for this system is defined as
Z = 1+e(µ−εA)/τ
where this represents the occupied states, where we know that the probability is given by
P(N1,ε1) =e(µ−εA)/τ
1+e(µ−εA)/τ
using the definition for the absolute activity
λ = eµ/τ
we can write the probability as
P(1,εA) =λe−εA/τ
1+λe−εA/τ = 0.9
thus we can see thatλ = 0.9(eεA/τ +λ)
which simplifies to
ln
(
λ0.9
−λ)
τ = εA = −13.71τ = −13.71×kBT
since we know thatT = 310 K thus we find that the energy is
εA = −.364 eV/O2
b) Now admit the CO under the specified conditions. FindεB such that only 10 percent of the Hb sitesare occupied by O2.
98
In this case we know that the grand partition function can be expressed as
Z = 1+λO2e−εA/τ +λCOe−εB/τ
where the first term is for an unoocupied state, the second term is when it is occupied byO2 and the thirdterm is for when it is occupied by CO. thus we can see that the probability is given by
P(N,εA) =λO2e
−εA/τ
1+λO2e−εA/τ +λCOe−εB/τ = 0.1
thus we find thatλO2e
−εA/τ = 0.1[1+λO2e−εA/τ +λCOe−εB/τ]
we can see that the energy is given by
εB = −τ ln
[[
λO2e−εA/τ
0.1−1−λO2e
−εA/τ
]
1λCO
]
using the energy found in part (a) we find that the energy is
εB = −0.543 eV/CO
Problem # 3 Concentration Fluctuations
The number of particles is not constant in a system in diffusive contact with a resevoir. We have seen that
〈N〉 =τζ
(
∂ζ∂µ
)
τ,V
from
〈N〉 =τζ
∂ζ∂µ
= τ∂ lnζ∂µ
a) Show that
〈N2〉 =τ2
ζ∂2ζ∂µ2
the mean-square deviation〈(∆N2)〉 of N from 〈N〉 is defined by
〈(∆N)2〉 = 〈(N−〈N〉)2〉= 〈N2〉−2〈N〉〈N〉+ 〈N〉2
= 〈N2〉−〈N〉2
〈(∆N)2〉 = τ2
[
1ζ
∂2ζ∂µ2 −
1ζ2
(
∂ζ∂µ
)2]
since we know that〈N2〉 = 〈(∆N)2〉+ 〈N〉2 (5.1)
99
we can see that
〈(∆N)2〉 = τ2
[
1ζ
∂2ζ∂µ2 −
1ζ2
(
∂ζ∂µ
)2]
and also we can see that
〈N〉2 =τ2
ζ2
∂2ζ∂µ2
putting this into equation 1 we find
〈N2〉 = τ2
[
1ζ
∂2ζ∂µ2 −
1ζ2
(
∂ζ∂µ
)2]
+τ2
ζ2
∂2ζ∂µ2
doing the math yields the solution we were looking for
〈N2〉 =τ2
ζ∂2ζ∂µ2
b) Show that this may be written as
〈(∆N)2〉 = τ∂〈N〉
∂µ
in Chapter 6 we apply this result to the ideal gas to find that
〈(∆N)2〉〈N〉2 =
1〈N〉
is the mean square fractional fluctuation in the population of an ideal gas in diffusive contact wit aresevoir. If〈N〉 is of the order of 1020 atoms, then the fractional fluctuation is exceedingly small. Insuch a system the number of particles is well defined even though it cannot be rigorously constantbecause diffusive contact is allowed with the resevoir. When 〈N〉 is low, this relation can be usedin the experimental determination of the molecular weight of large molecules such as DNA of MW108−1010.
Since we know that
〈N〉 =τζ
∂ζ∂µ
where we can see that∂〈N〉
∂µ=
∂∂µ
(
τζ
∂ζ∂µ
)
using the product rule we find∂〈N〉
∂µ=
τζ
∂2ζ∂µ2 −
τζ2
∂2ζ∂µ2
thus we find that
〈(∆N)2〉 = τ2[
τζ
∂2ζ∂µ2 −
τζ2
∂2ζ∂µ2
]
thus we have just shown that
〈(∆N)2〉 = τ∂〈N〉
∂µ= τ2
[
τζ
∂2ζ∂µ2 −
τζ2
∂2ζ∂µ2
]
100
Problem # 4 Multiple binding of O2
A hemoglobin molecule can bind four O2 molecules. Assume theε is the energy of each bound O2,relative to O2 at rest at infinite distance. Letλ denote the absolute activityeµ/τ of the free O2 (in solution).
a) What is the probability that one and only one O2 is adsorbed on a hemoglobin molecule? Sketch theresult qualitatively as a function ofλ.
We know that the grand partition function for this system is given by
Z = ∑ASN
e(Nµ−εs(N))/τ = 1+4e(µ−ε)/τ +6e2(µ−ε)/τ +4e3(µ−ε)/τ +e4(µ−ε)/τ
This is because we know that if there are no molecules attached then we get the first term, the second termis assuming that there is one oxygen atom attached we find the multiplicity to be 4, if there are two atomsattached then the multiplicity is 6, if there are 3 atoms attached then the multiplicity is 4, and if there arefour atoms attached then the multiplicity is 1. Since we knowthat the absolute activity is defined as
λ = eµ/τ
allows us to write the grand partition function as
Z = 1+4λe−ε/τ +6λ2e−ε/τ +4λ3e−ε/τ +λ4e−ε/τ
we know that the probability that one atome is attached is given by
P(N = 1,ε) =4λe−ε/τ
1+4λe−ε/τ +6λ2e−ε/τ +4λ3e−ε/τ +λ4e−ε/τ =4λ
eε/τ +4λ+6λ2+4λ3+λ4
The plot is given
0 200 400 600 800 1000 1200λ
0.00
0.01
0.02
0.03
0.04
P(λ
)
b) What is the probability that four and only four O2 are adsorbed? Sketch this result also.
101
This probability is given by
P(N = 4,ε) =λ4
eε/τ +4λ+6λ2+4λ3+λ4
the plot is given by
102
Chapter 6
The Ideal Gas
Roadmap
• Identical Particles: Bosons and Fermions
• Fermi-Dirac Distribution
• Bose-Einstein Distribution
• Classical Limit and the Chemical Potential
• Ideal Classical Gas,F, p,µ,σ,CV
• Expansion of Gases
• Polytomic gases: Internal degrees of freedom and Equipartitio Theorem
6.1 Identical Particles: Bosons and Fermions
Lets start of by considering a three dimensional space with two identical particles. This are particlesthat cannot be labeled, they are strictly identical. We can write down a Shrodinger equation that is timeindependent
− ~2
2m∇2
1ψT − ~2
2m∇2
2ψT +(V1+V2)ψT = ETψT
we are assuming that this particles are i some electric field that give them the potential energy. We cansolve by writing down the wave functios by bte product of the two wave functionc
ψT = ψα(1)ψβ(2)
whereα andβ are te quantum numbers of each particle. If we plug this in we get the following result
− ~2
2mψβ(2)∇2
1ψα(1)− ~2
2mψα(2)∇2
2ψβ(2)+(V1+V2)ψα(1)ψβ(2) = ETψα(1)ψβ(2)
we can see that this two equations seperate. This can be written as
− ~2
2m
∇21ψα(1)
ψα(1)+V1 = E1 − ~
2
2m
∇22ψβ(2)
ψβ(2)+V2 = E2 ET −E1+E2
103
this is for non-interacting particles only. So far so good, we found a wave function that satisfy the SE.This are identical particles and QM will not allow us to labelthem, it only allows us to distinguish theirstates. We cannot label identical particles, so an equaly good wave function is of the formψα(2)ψβ(1) sothe general solution is of the form
ψT =1√2[ψα(1)ψβ(2)±ψβ(1)ψα(2)]
this brings us to the question of the symmetry of the wave function, symmetric and anti-symmetric. Theconsequencies of this is very profound. What we learn emperical is that ther are two general classes ofbehaviours. The first is for fermions. Fermions have a 1/2 integer spins, some examples are electrons,protons,neutrons, and helium 3. All of these have 1/2 spin. The other class are the bosons with integerspins, i.e photons, deuterons, and helium 4 nucleus. So the first class, the fermions have anit-symmetricwave functions and the bosons have symmetric wave functions. What difference does this make? Ifwe interchange the symmetric wave functions we can see that there is no change, on the other handinterchanging the anti-symmetric wave functions we can seethat there is a changeψ− → −ψ−. Whathappens if we letα = β. For the symmetric case, there is nothing that is violated, the total wave functionsimply becomes
ψT =2√2
ψα(1)ψα(2)
if we do this for the anti-symmetric case we see that the wave function vanishes. This is a result of thePauli exclusion principle, you cannot have two identical particles in the same quantum state. There is norestriction for bosons.
6.2 Fermi-Dirac Distribution
We will use the grand canonical ensemble, we must consider a alot of particles in discrete quantu levels.We will pick out 1 state that we will call the system and the rest we will consider the resevoir. Our goal isto find the average occupancy that has an energyεl . Since these are fermios means that we can only have1 particle in a given state. We know thatN = 0 or N = 1. This system is in thermal equilibriums that canexchange energy and particles. We want to find the average number of particles in this system.
〈N(εl)〉
we must write down the grand partition function.
ζ = 1+e(µ−εl )/τ
these are the only two allowed states
〈N(εl)〉 =e(µ−εl )/τ
1+e(µ−εl )/τ
this is know asf (ε)
f (ε) =1
e(ε−µ)/τ +1
this is known as theFermi-Dirac distribution function. We can see that this result involvesthe potentialenergy. This formula has tow different interpretations, one is that the probability that the state with energy
104
ε is occupied, or it is the averagge occupancy of state with energy ε. What is the form of this distribution?What will this look like atτ = 0? If we take energy
ε < µ e(ε−µ)/τ = e−∞ = 0 f (ε < µ) = 1
ε > µ e(ε−µ)/τ = e∞ = ∞ f (ε < µ) = 0
if we were to plot this we would get a step function atµ. Secondly, at any temperature
f (ε = µ) =1
e0+1=
12
thusµ is the energy for which the occupancy is 1/2. What happens with increasing temperature? Letssuppose that we have a 3-D gas with a Fermi gas. Lets take spin 1/2 with a lot of energy levels and wealso have an even number of Fermions. The lowest energy levelwill have only two electrons (exclusionprinciple). The next two will go into the first excited state etc. We will see that the electrons in the higherenergy state will move to even higher energy states. We will smear out the distribution around the energylevelsτ(kBT). The Fermi level isεF , and atT = 0 it is exactly the same as the chemical potentialµ. Thissmearing is symmetric around the chemical potential.
Symmetry of Filled and Vacant states
What is the probability that the stateµ= δ is
f (µ−δ) =1
e−δ/τ +1f (µ+δ) =
1
eδ/τ +1
this is the probability that these two states will be occupied. What is the probability that atµ+δ is vacant?
p(µ+δ) = 1− f (µ+δ) =(eδ/τ +1)−1
eδ/τ +1=
eδ/τ
eδ/τ +1=
1
e−δ/τ +1= f (µ−δ)
if we now imagine what happens to this distribution as we increase the temperature. We can see that therewill be a smearing of the dsitribution that is of the order ofτ. The chemical potential will eventually startto change, asτ is increased thanµ drops.
6.3 Bose-Einstein Distribution
This applies to bosons (particles with integer spins), where the Pauli exclusion principle does not apply.Once again we must consider a resevoir and a system. We can putas many bosons as we like in anyquantum state. As before we consider 1 quantum statel that is in thermal equilibrium with all the otherstates in the resevoir. We ca now write down the grand partition function
ξ = 1+e(µ−εl )/τ +2e2(µ−εl )/τ + ...
bothµ andε get multiplied by the number
ξ =N0→∞
∑n=0
en(µ−εl )/τ
105
we will assume thatN0 → ∞. If we now let
x = e(µ−εl )/τ
thus we find
ξ =∞
∑n=0
xn =1
1−x
this only works ifx < 1, otherwise this function blows up. We can see that
µ− εl
τ< 0 µ< ε
what we now want to calculate the average occupancy of the state l .
〈n(ε)〉 6= probability of state being occupied
this is given as
〈n(ε)〉=∑nxn
∑xn =x d
dx ∑xn
∑xn =x d
dx
(
11−x
)
11−x
=
x(1−x)2
11−x
=1
x−1
thus we find the Bose-Einstein distribution
〈n(ε)〉= n(ε) =1
e(ε−µ)/τ −1
the Bose-Einstein distribution will always haveε > µ. The Bose-Einstein distribution blows up ase(ε−µ)/τ →1, this function converges the the Fermi-Dirac distribution whenε−µ≫ τ. The Bose-Einstein condensa-tion refers to the the fact that at low enough temperatures all of the bosons will condense into the groundstate.
6.4 The Classical Limit and the Chemical Potential
We would like to recap the 1 particle in the box solutions. We derived the quantum concentration
nQ =( mτ
2π~2
)3/2≈ 1
λ3D
thus for 2 particles seperated by more than the Debroigle wavelength, we find that the concentration ofthe particlesn≈ nQ. If we have two particles that are very close with respect to the Debroigle wavelengthwe must consider the quantum effects and must utilize quantum statistics. If we have two particles thatare seperated by much more than the Debroile wavelength thanwe will be in the “classical limit”n≪ nQ.These scales are set by the Debroigle wavelength. In the context of 1 particle we found
Z1 = ∑n=0
e−εn/τ = nQV
wheren2 = n2x + n2
y + n2z andV is the volume of the box. Once again we must remember that we are in
the tail of the distribution where the average occupancy is much less than 1. We define the calssical gas isone in whichf (ε) or n(ε) is≪ 1. This is a low density level, at this level, both distributions give the sameresults. So that
e(ε−µ)/τ ≫ 1
106
thus we findn(ε) = eµ/τ ·e−ε/τ
this is the distribution function for a classical gas. We want to now figure something out about the chemicalpotential. Lets confine our attention to a monotomic gas, i.eHe. What isµ? We can findµ from thefollowing
N = 〈N〉 = ∑n
f (εn)
we simply add the occupancies of all the states. We find
N = eµ/τ ∑n
e−εn/τ = eµ/τZ1 = nQVeµ/τ
this is rather neat, we can devide by the volume on both sides to get
NV
= nQeµ/τ ⇒ nnQ
= eµ/τ
taking logs on both sides
µ= τ ln
(
nnQ
)
this links the classical potentialµ to the densities. If we we look at our previous expression
n(ε) = f (ε) = eµ/τe−ε/τ =
(
nnQ
)
e−ε/τ
thus f (ε) ≪ 1 providedn ≪ nQ. We can see from the expression of the chemical potential that for aclassical gas, the chemicla potential is negative. This is yet another signature between the classical gasand the quantum gas. If we lowern/nQ we can see that the chemical potential changes signs atn = nQ,where we go from classical to quantum regimes. The definitionof the chemical potential always involvesnQ which always involve~. Thus even at the classical limit, the particles “remember” that they are madeup of indistinguishable particles.
Example: Variation of the Pressure of the Earth’s Atmosphere with Height
We know the chemical potential is given by two quantities
µtot = µint +µext
we will make a an assumption thatτ is fixed. If we consider two layers of gas that are in diffusivecontact,which allows particles and energy to be transferrable through the layer. The layer is filled with monotomicparticles in thermal equilibrium. If we also assume that we have zero potential energy at sea level. We cansee that the particles above sea level will have
µext = mgd
thus the total chemical potential is
µtot = τ ln
(
nnQ
)
+mgd
107
now, in thermal equilibrium, the total chemical potential must be constant. We can see that as we go upin altitude,µext increases andµint will decrease. We can make the following statement: if we consider theupper level
τ ln
(
n(d)
nQ
)
+mgd= τ ln
(
n(0)
nQ
)
combining the two log terms we find
τ ln
(
n(d)
n(0)
)
= −mgd
that is to say thatn(d) = n(0)e−mgd/τ
we know that at a given temperature, the pressure will be proportional to
p(d) = p(0)e−mgd/τ = p(0)e−d/d(0)
whered(0) = τ/mg, if we put some numbers, i,.eN2 we find
d(0) =290 K
28×1.7×10−27 kg×9.8 m/s2≈ 8.7
6.5 Ideal Classical Gas:Free Energy, Pressure, Internal Energy, En-tropy, and Heat Capacity
The word ideal means that there are no interaction between the particles, and classical mean that we arein the low energy limit. The free energy is the most importantquantity one will derive. Recall that thechemical potentialµ is
µ= τ ln(n/nQ) nQ =
(
Mτ2π~2
)3/2
the first thing we need to find is the free energy
6.5.1 Free Energy
µ=
(
∂F∂N
)
τ,µ
where
µ= τ[
lnN− lnV − 32
lnτ− 32
lnM
2π~2
]
this is for later convenience, so we can then write
F(N,τ,V) =
Z N
0dNµ(N,τ,V) = τ
Z N
0dN
[
lnN− lnV − 32
lnτ+32
ln
(
2π~2
M
)]
lets recall the following resultZ
lnxdx= xlnx−x
108
we get the following
F(N,τ,V) = τ[
N lnN−N−N lnV − 3N2
lnτ+3N2
ln
(
2π~2
M
)]
(6.1)
the first two terms give youN lnn = N lnN−N−N lnV
and the last two terms give you-
−N lnnQ =3N2
lnτ+3N2
ln
(
2π~2
M
)
thusF(N,τ,V) = Nτ lnn−Nτ lnnQ = Nτ[lnn− lnnQ]
thus
F(N,τ,V) = Nτ[
ln
(
nnQ
)
−1
]
this is the free energy.
6.5.2 Pressure
The pressure is defined as
p = −(
∂F∂V
)
τ,N
using Equation 6.1 we find
p = −(
−NτV
)
thus we get the ideal gas lawpV = Nτ = NkBT = RT
whereR is the gas constant for 1 mole. We can immediatly figure out that
R= NAkB = 6.02×1023atoms/mole×1.38×10−23J/K = 8.31 J/K mole
6.5.3 Internal Energy
Lets write down the thermodynamic identity
dF = −σdτ− pdV+µdN
this enables us to figure out the free energy very easily
F = u− τσ u = F + τσ σ = −(
∂F∂τ
)
V,N
thus the internal energy is
u = F − τ(
∂F∂τ
)
109
there is a trick, lets consider
−τ2[
∂∂τ
(
Fτ
)]
= −τ2
(
− 1τ2F +
(
∂F∂τ
)
V,Nτ
)
= F − τ(
∂F∂τ
)
thus
u = −τ2[
∂∂τ
(
Fτ
)]
V,N
but from Equation 6.1 we can see that there is only one term inτ, thus
u = −τ2(
−3N2
)
1τ
=32
Nτ
which is the same result we have derived before from the equipartition theory.
u =32
Nτ
Note: The internal energy of an ideal gas only depends on the temperature.
6.5.4 Entropy
We can go back to the thermodynamic identity, we find that the entropy is given by
σ = −(
∂F∂τ
)
V,N
once again looking at Equation 6.1, and differentiating with respect toτ, hence
σ = −[
N lnN−N−N lnV − 32
Nτ+32
N ln
(
2π~2
M
)
− τ3N2
1τ
]
where the last term came from∂∂τ
(
3N2
lnτ)
thus we can see that the entropy is given by
σ = −[
N lnn− 32
N ln(τ)+32
N ln
(
2π~2
M
)]
+52
N
we also know that
− ln(nQ)N = −32
Nτ+32
N ln
(
2π~2
M
)
putting all these together we have
σ = N
[
ln(nQ
n
)
+52
]
this result is called theSackus-Tetrode Equation.This is the central result of the ideal gas, which willallow you to derive all the other quantities. Note: Gibbs paradox (∼ 1875) was the first indication ofproblems with classical physics (indistinguishibility)
110
6.5.5 Heat Capacity
We havedu= τdσ− pdV
where the first term is the heat added and the second term is thework done on the gas. There are two kindsof heat capacity
• Heat capacity at constant volume. We have a box with a bunsen burner, the heat increases but thevolume stays the same
• Heat capacity at constant pressure. We have a piston with pressure being applied, along with abunsen burner that adds heat to the system.
In the first case there is no mechanical work, and in the secondcase there is mechanical work that isapplied in pushing back the piston against constant pressure. We can see that
CP > CV
lets start withCV . If we want to increase the temperaturedτ then the internal energy must increase bydu.The internal energy depends only on the temperature. We can see
(
δuδτ
)
V= τ(
δσδτ
)
V
thus
CV =
(
∂u∂τ
)
V= τ(
∂σ∂τ
)
V
we rember that the entropy is
σ = N
[
lnnQ− lnn+52
]
= N
(
32
lnτ+ terms independent ofτ)
where we can see that(
∂σ∂τ
)
=32
N1τ
thus the heat capacity is
CV =32
N =32
NkB
now for the heat capacity at constant pressureCP, we begin with
τdσ = du+ pdV
thus
CP = τ(
∂σ∂τ
)
P
we can now write this as follows
CP =
(
∂u∂τ
)
P+ p
(
∂V∂τ
)
111
the first term is once againCV , that is because the internal energy depends only on the temperature. Thesecond term is the work done in expanding the gas, which we must figure out, thus
Cp = CV + p
(
∂V∂τ
)
P
thus we can see that
pV = Nτ V =Nτp
thus
p
(
∂V∂τ
)
P= p
Np
= N
thus
CP = CV +N =52
N =52
NkB
and we can see thatCP > CV
this is only for a monotomic gas. The final thing is to define theratio ofCP toCv where
γ =CP
CV=
53
6.6 Expansion of Gases
There are two kinds of expansion in gases, one in which we keepthe temperature fixed and the other wherewe keep the pressure fixed.
6.6.1 Isothermal Quasistatic Expansion
Quasistatic implies slow which implies that this is in thermal equilibrium during expansion. We have someresevoir that we keep at constant temperatureT0, we then have some cylinder with a piston that applies aconstant pressure on the system. What we know is thatpV = Nτ and since we know that the temperatureis constant we know thatpiVi = pfVf , this is known asBoyle’s Law. What we know is
σ = N
[
ln(nQ
n
)
+52
]
= N lnV + terms independent ofV
in this context the entropy only depends on the first term. Thechange in the entropy is given as
∆σgas= N lnVf −N lnVi = N ln
(
Vf
Vi
)
if the final volume is bigger then the initial volume then we can see that the entropy increases. What is thework doneon the gas?
∆W = −Z Vf
Vi
Fdx= −Z Vf
Vi
FA
dV = −Z Vf
Vi
pdV
going back to the ideal gas law we find
∆W = −Z Vf
Vi
NτV
dV = −Nτ ln
(
Vf
Vi
)
112
apart from the−τ this is the same as the entropy, thus
∆W = −τ∆σgas
and finally the change in the internal energy∆u = 0
because the temperature of the gas did not change. We could have guessed this result if we think about theinternal energy
u =32
Nτ
thus∆u = 0
from this we can talk about the heat flow from or into the gas, ifthe change in the internal energy is zerowe see right away that we have the following result
∆Qgas+∆W = 0
or∆Qgas= −∆W
if we now do work to push up the piston, we can see that we will dowork to make this happen. This ispart of the thermal cycle of a heat engine. This is the heat flowduring expansion. The final thing we canask is what is the change i the entropy of the resevoir, this isgiven by
change in entropy of resevoir=∆Qgas
τ= −∆Qgas
τ=
∆Wτ
= −∆σgas
and we can see that∆σres+∆σgas= 0
which implies∆σ = 0
this implies that this is an entirely reversible process. Wewill now turn to the adiabatic expansion.
6.6.2 Adiabatic Quasistatic Expansion
The word adiabatic is tricky, in the case of QM this is a very slow process, in the case of thermodynamicsthe word adiabatic means that it is thermally isolated. Conceptually, we have a cylinder and a piston withgas atoms inside, but there is no heat exchange with the system and the resevoir. When we change thevolumeVi →Vf there is no heat exchange (∆Q = 0 and∆σgas= 0) with the environment and thus this isan adiabatic process. We want to find a relationship between the pressure and the volume. Lets figure outwhat this is, we will end up usingCP,CV andγ = CP/CV . We will begin using the thermodynamic identity
du= τdσ− pdV
we know thatdσ = 0 and we also know
CV =
(
∂u∂τ
)
Vdu= CVdτ
113
and we can see thatCVdτ+ pdV = 0
we need to eliminatedτ, we can use the ideal gas equation
d(pV) = d(Nτ)pdV+Vdp = Ndτ
we can seeCV(pdV+Vdp)+NpdV= 0
we can rewrite this as followspdV(CV +N)+CVVdp= 0
we can see that this simplifies to
CP
CVpdV+Vdp = 0
γpdV+Vdp = 0
γdVV
+dpp
= 0
γ lnV + ln p = C
which givespVγ = Constant
this is the result used for adiabatic expansions. Also, letsconsider
pV = Nτ p =NV
τ
we can see thatτVγ−1 = Constant
or
Vγ =Nγτγ
pγ
which yieldspγ−1τγ = Constant
the last thing we should work out is the work done on the gas. Weuse
∆W = uf −ui
this is simply
∆W =32
N(τ f − τi)
=32
Nτi
(
τ f
τi−1
)
=32
Nτi
(
(
Vf
Vi
)1−γ−1
)
114
which gives
∆W =32
Nτi
[
(
Vf
Vi
)−2/3
−1
]
when the gas expands there is a increase inσ due to the increase in the volume. If the temperature drops,this gives a decrease inσ which exactly cancels the increase inσ due to the volume increase.
Comment on “adiabatic” process in QM
lets imagine that we have many energy levels in a box that are populated with gas atoms. We will expandthe box very slowly so that the energy levels are all lower. The result of doing time dependent pertubationtells us that the atoms do not change energy levels, and thus the distribution of the particles over thequantum states on the average does not change. The consequence is that the entropy is constant.
6.6.3 Sudden Expansion in a Vacuum
Once again this is an isolated system. Lets imagine that we have a box with a partition in it. Initially allthe atoms are confined to the left hand side, we than remove thepartition very suddenly. We thus go into anew state in whihc the particles are distributed uniformly over the volume of the box. We can see that thisis an irreversible process, i.e placing the partition back in will not give us the same distribution. First ofall, there is no work done during this process, and thus the energy of the system does not change∆u = 0,this is because the internal energy only depends on the temperature. By the same token∆τ = 0, but thechange in the entropy is not zero, it is
∆σ = N ln
(
Vf
Vi
)
this is analogous to what happens in QM, we can compare this tothe sudden approximation in QM, i.ebeta decay which gives rise to a new set of atomic states.
6.7 Polyatomic Gases: Internal Degree of Freedom
Example: N2 (Dumbell)The first thing is that we can have vibration and the second thing is that they can rotate. This gives rise
to significant changes in the heat capacity. If we think of this classically we will need energykT to excitethe vibration. We will imagine a set of distinct energy levels given by
ε =12
n~ω n = 1,3,5, ...
in the classical limit we know thatu = kBT thus the heat capacity isCV = kB, this of course is in the limitkBT ≫ ~ω. What about rotation? There are two degrees of rotation for polyatomic gases, thus the energyof a rotator is given as
urot =12
Iω2
whereI is the moment of inertia, there is nonly a kinetic energy term. We can also write this as
urot =12
(Iω)2
I
115
in QM we have the following〈(Iω)2〉 = r(r +1)~2
wherer is an integer andIω is the angular momentum. Thus the rotational energy is givenby
urot =r(r +1)~2
2Ir = 0,1,2,3, ...
once again since this is in the classical limit we will have12kBT for each degree of rotation, thus
urot(2 modes) = kBT
and the heat capacity will beCV = kB
the rotation mode about the axis through the atoms is never excited, it would take an infinite amount ofenergy since the moment of inertia is approximately 0. If we were to go toNH3 we would have 3 modesof rotation. Given this result we can now ask how big is the rotational energy.
Rotational energy ofN2
Lets consider the seperation between the center of rotationand one atome be 10−10 meters. What is themoment of inertia, we know thatMN2 ≈ 2×10−26kg, we find the moment of inertia to by
I = 2× (10−10)2×2×10−26kg≈ 4×10−26m2kg
thus the energy of the first excited state
u =2× (10−34)2
2×4×10−26 ≈ 2×10−22J≈ 10 K
this gives a reasonable temperature above which you would get rotation.
6.7.1 The Grand Partition Function for the ideal Gas with Internal Degrees ofFreedom
This is used when the problem implies that there are internaldegrees of freedom. The grand sum forbosons and fermions become the same in the classical limit. For he case o the fermions when you havetwo energy levels you can have at most one particles in the ground state. For fermions
ζ = 1+e(µ−εn)/τ
for bosons we haveζ = 1+e(µ−εn)/τ +e2(µ−εn)/τ + ...
in the case of a classical gas we can see that the higher order terms are negligable and that the grandpartition functions are the same. Sincef (εn) ≫ 1 we can neglect the higher order terms. Hence in theclassical gas we take as the grand partition function
ζ = 1+e(µ−εn)/τ
116
what do we do if we have additional terms? We must add another term that describes internal degrees offreedom. For the polyatomic caseεn → εn+ εint , where this corresponds to the vibrational and rotationalenergy, thus we now have
ζ1+∑int
e(µ−εn−εint)/τ
thus we must sum over all the possible internal degrees of freedom. We can now write this in the form
ζ = 1+e(µ−εn)/τ ∑e−εint/τ
we will call the multiplying factorZint , which yields
ζ = 1+Zinte(µ−εn)/τ
we can use this to find a new expression for the chemical potential. The probability that the staten isoccupied for any state of its internal motion is given by the following
f (εn) =Zinte(µ−εn)/τ
1+Zinte(µ−εn)/τ
but now by hypothesis we have agreed that the grand partitionfunction is very small, thus we find
f (εn) = Zinte(µ−εn)/τ
this is the result for the monotomic gas multiplied by the factor representing the internal degrees of free-dom. We can now relate this to the average occupancies
N = 〈N〉 = ∑ f (εn) = Zinteµ/τ ∑e−εn/τ = Zinte
µ/τZ1
which givesN = Z1Zinte
µ/τ
but we remember thatZ1 = nQV
we can now writeN = nQVZinte
µ/τ
rearranging gives us the followingn
nQ= Zinte
µ/τ
taking the log of both sides give
µ= τ[
ln
(
nnQ
)
− lnZint
]
we can see that we now have an additional term that accounts for the internal degrees of freedom, withoutthis term we simply recover the result for the monotomic gas.
117
6.7.1.1 Example 1 (Diesel Engine)
This is a process in which gas is compressed adiabatically toa high enough temperature to ignite the dieselfuel that is then injected. If we start we air atT = 300 K and lets suppose that the compression ratio is 16,which means that we reduce the volume to 16 times smaller. We have seen that we get the following result
TiVγ−1i = TfV
γ−1f
thus
Tf = Ti
(
Vi
Vf
)γ−1
we know thatγ = 7/5 for air, which is nitrogen and oxygen, this is because
CV =32
NkB +NkB
where the second term is added for the rotational mode, and weknow that
CP = CV +NkB =72
NkB
and
γ =75
if we put the number in we getTf = 300(16)2/5 = 909 K
6.7.1.2 Example 2
We need to show that
p =32
UV
this is done in three steps.
a) Find that the average pressure in a system in thermal contactwith a heat resevior. Recall
ps = −(
∂εs
∂V
)
N
thus the average pressure is equal to〈p〉 = ∑
spsP(εs)
wherep is the pressure andP is the probability that states is occupied. We find
〈p〉 = −∑s
(
∂εs
∂V
)
N
e−εs/τ
Z
this is step 1.
118
b) Show that for an ideal gas(
∂εs
∂V
)
N= −2
3εs
V
there are various ways of getting this. If we have a cube of sideL we find
εs =~
2
2mπ2
L2(n2x +n2
y +n2z) ∝
1
V2/3
for all nx,ny, andnz. If we take logs of both sides we find
lnεs = −23
lnV
if we differentiate both sides we finddεs
εs= −2
3dVV
which gives(
∂εs
∂V
)
N= −2
3εs
V
c) the result of the pressure
p = −1Z ∑
s
(
−23
εs
V
)
e−εs/τ =2
3V
(
∑sεse−εs/τ
Z
)
=2U3V
if we go for a monotomic gas, the energyU = 32NkBT we can see that
pV = NkBT
which is the ideal gas law.
6.8 Summary
The first thing we learned is
f (ε) =1
e(ε−µ)/τ ±1
where±1 account for either a boson or fermion. In the classical limit ε−µ≫ τ we can see that
f (ε) = eµ/τe−ε/τ = λe−ε/τ
whereλ is the absolute acitivity. we also found
µ= τ ln(n/nQ)
alsoF = Nτ[ln(n/nQ)−1]
andpV = Nτ
119
also the internal energy
U =32
Nτ
and the entropy is
σ = N
[
ln(n/nQ)+52
]
the heat capacity is
CV =32
N
and
Cp = CV +N =52
N
we also learned about isothermal expansion
piVi = pfVf
we also saw that∆σres = −N ln(Vf /Vi) ∆σgas= N ln(Vf /Vi)
which gives∆σ = 0
we also talked about the adiabatic expansion
piVγi = pfV
γf γ =
CP
CV
where this process is a reversible process. We also talked about the free expansion, this is where we rupturea membrane in some box. We found that there is no change in the internal energy and no change in thetemperature
∆U = 0 ∆τ = 0 ∆σ = N ln(Vf /Vi)
this process is irreversible. We also talked about diatomicgases. We found
CV =32
N+N+N
where these are for rotational, vibrational and translational energies. We also find
µ= τ [ln(n/nQ)− lnZint ]
we can also remember thatthe ideal gas does not exist.
6.9 Problems and Solutions
Problem # 1 Distribution Function for Double Occupancy Statistics
Let us imagine a new mechanics in which the allowed occupancies of an orbital are 0,1, and 2. Thevalues of the energy associated with these occupancies are assumed to be 0,ε , and 2ε, respectively.
a) Derive an expression for the ensemble average occupancy〈N〉, when the system composed of thisorbital is in thermal and diffusive contact with a reservoirat temperatureτ and chemical potentialµ.
120
We know that the average occupancy is defined as
〈N〉 =∑N ∑sNe(Nµ−εs)/τ
ζ
whereζ is the grand partition function
ζ = ∑N
∑s
e(Nµ−εs)/τ (6.2)
the average occupancy can also be acquired witha bit of algebreac manipulation, Consider
∂ζ∂µ
=Nτ
ζ =1τ
(
∑N
∑l
N(Nµ−εl )/τ
)
=ζ〈N〉
τ
what we see is that the average value ofN is given by
〈N〉 =τζ
∂ζ∂µ
= τ∂ lnζ∂µ
thus we simply need to know what the grand sum is. This is givenby Equation 1, we find (if we letλ = eµ/τ)
ζ = 1+λe−ε/τ +λ2e−2ε/τ
where the derivative is given as
∂ lnζ∂µ
=1ζ
∂ζ∂µ
=1ζ
[
1τ
λe−ε/τ +2τ
λ2e−2ε/τ]
thus we can see that the average occupnacy is given as
〈N〉 =λe−ε/τ +2λ2e−2ε/τ
1+λe−ε/τ +λ2e−2ε/τ
b) Return now to the usual quantum mechanics, and derive an expression for the ensemble average occu-pancy of an energy level which is doubly degenerate; that is,two orbitals have the identical energyε. If both both orbitals are occupied the total energy is 2ε.
We can write the grand partition function for this system, ifwe know thats= 0, εs = 0 ands= 1, εs = εand this is doubly degenerate, and for the last orbitals= 2, εs = 2ε allows us to write the grand sum as
ζ = 1+2λe−ε/τ +λ2e−2ε/τ
and the ensemble average is given by
〈N〉 = 2
[
λe−ε/τ +λ2e−2ε/τ
1+2λe−ε/τ +λ2e−2ε/τ
]
Problem # 2 Energy of Gas of Extreme Relativistic Particles
121
Extreme relativistic particles have momentump such thatpc≫ Mc2, whereM is the rest mass of theparticle. The de Broglie relationλ = h/p for the quantum wavelength continues to apply. Show that themean energy per particle of an extreme relativistic ideal gas is 3τ if ε ≃ pc, in contrast to3
2τ for thenonrelativistic problem.
We know that
λ =hp
ε = pc ε =hcλ
but we know that
L =n2
λ λ =2Ln
thus
ε(n) =hcn2L
We know that the mean energy is given as
U = τ2∂ lnZ∂τ
thus we must find the partition function, which is given as
Z = ∑n
e−εn/τ =π2
Z ∞
0n2e−hcn/2Lτdn
if we let
x =hcn2Lτ
dn=2Lτhc
dn n2 = x2(
2Lτhc
)2
thus we find that the partitio function is given by
Z =
(
2Lτhc
)3 π2
Z ∞
0x2e−xdx
whereZ ∞
0x2e−xdx=
[
−x2e−x−2xe−x−2e−x]∞0 = 2
thus we find that the partition function is
Z = π(
2Lτhc
)3 ∂Z∂τ
= 3πτ2(
2Lτhc
)2
thus we find
U =τ2
Z∂Z∂τ
=3πτ4
πτ3 = 3τ
Problem # 3 Entropy of Mixing
122
Suppose that a system ofN atoms of type A is placed in diffusive contact with a system ofN atoms oftype B at the same temperature and volume. Show that after diffusive equilibrium is reached the totalentropy is increased by 2N log2. The entropy increase 2N log2 is known as the entropy of mixing. If theatoms are identical (A≡ B), show that there is no increase in entropy when diffusive contact isestablished. The difference in the results has been called the Gibbs paradox.
We know that the entropy of an ideal gas is given by
σ = N
[
ln(nQ
n
)
+52
]
and we also know that the total entropy before the mixing is just the sum of the individual entropieswhich is given by
σT = σA +σB
this is simply
σT = NA
[
ln
(
nQ
nA
)
+52
]
+NB
[
ln
(
nQ
nA
)
+52
]
we also know
N = NA = NB nA =NA
VnB =
NB
Vn = nA = nB
thus we find that the total entropy is given by
σT = 2N
[
ln(nQ
n
)
+52
]
the entropy after mixing can be found by assuming
NT = 2N n=NV
nf = 2n
thus the entropy is now given as
σmix = 2N
[
ln(nQ
2NV)
+52
]
and we find that the change in the entropy is given as
∆σ = σT −σmix = 2N
[
ln(nQ
n
)
+52
]
−2N
[
ln(nQ
2n
)
+52
]
thus we can see that this is simply∆σ = 2N ln2
Problem # 4 Time for a Large Fluctuation
We quoted Boltzmann to the effect that two gases in a 0.1 litercontainer will unmix only in a timeenormously long compared to 10(1010) years. We shall investigate a related problem: We let a gas ofatoms of4He occupy a container of volume of 0.1 liter at 300 K and a pressure of 1 atm, and we ask howlong it will be before the atoms assume a configuration in which all are in one-half of the container.
a) Estimate the number of states accesible to the system in thisinitial condition.
123
We know that the number of accesible states is given by
σ = lng g= eσ
where the entropy is defined as
σ = N
[
ln(nQ
n
)
+52
]
where the quantum concentration is given as
nQ =
(
Mτ2π~2
)3/2
n =NV
thus we find
σ = N
[
lnnQ− lnn+52
]
= N
[
32
ln
(
Mτ2π~2
)
− ln
(
NV
)
+52
]
where we also know
N =pVkBT
M = 4amu= 4×1.66×10−27 kg 0.1L = 10−4 m3 1 atm= 1.01×105 Pa
thus we find the entropy to be given as
σ =pVkBT
[
32
ln
(
MkBT2π~2
)
− ln
(
pkBT
)
+52
]
= 3.69×1022
and the number of accesible states is given by
g = e3.69×1022
b) The gas is compressed isothermally to a volume of 0.05 liter.How many states are accessible now?
We need to find what the change in the entropy is for this, this is given by
σ2 = N ln
(
V2
V1
)
+σ1 = N ln
(
12
)
+σ1
where these becomes
σ2 =pVkBT
ln
(
12
)
+σ1 = 3.52×1022
thus we find that the number of accesible states is now given by
g2 = e3.52×1022
c) For the system in the 0.1 liter container, estimate the valueof the ratio
number of states for which all atoms are in one-half of the volumenumber of states for which the atoms are anywhere in the volume
We know that this is simply
ratio=g2
g= e−1.70×1021
124
d) If the collision rate of an atom is≈ 1010s−1, what is the total number of collisions of all atoms in thesystem in a year? We use this as a crude estimate of the frequency with which the state of the systemchanges.
We know that the total number of collisions of all atoms is given by
total collisions for all atoms= Nr = 2.415×1021×1010(
π×107 syr
)
= 7.58×1038collisionsyr
e) Estimate the number of years you would expect to wait before all atoms are in one-half of the volume,starting from the equilibrium configuration.
We can do this by knowing that the ratio we calculated inc gives us the probability that all of the atomswill be in half of the container, and we can see that the time isgiven by
t = (p f)−1 =1p f
wherep is the probability that the atoms will be in 1/2 of the container andf is the frequency we found inpartd, so we find that the time is
t =e1.7×1021
7.58×1038yrs
Problem # 5 Gas of atoms with internal degree of freedom
Consider an ideal monotomic gas, but one for which the atom has two internal energy states, one energy∆ above the other. There areN atoms in volumeV at temperatureτ. Find the (a) chemical potential; (b)free energy; (c) entropy; (d) pressure; (e) heat capacity atconstant pressure.
a) Chemical potential?
We know that the chemical potential is found from the condition that the thermal average of the totalnumber of atoms equals the number of atoms known to be present. This number must be the sum over allorbitals of the distribution functionf (εs)
N = 〈N〉 = ∑s
f (εs)
The Gibbs sum for this system is given as
ζ = 1+λe−εn/τ = 1+λ∑int
e−(εn+εint)/τ = 1+λZinte−εn/τ
whereZint = ∑
inte−εint/τ
The probabiliy that the translational orbitaln is occupied is given by the ratio of the term inλ to the Gibbssumζ
f (εn) =λZinte−εn/τ
1+λZinte−εn/τ ≃ λZinte−εn/τ
125
if we assume thatf (εn) ≪ 1. Thus we can see that
N = λZint ∑n
e−εn/τ (6.3)
whereZint = ∑
inte−εint/τ = 1+e−∆ε/τ ε1 = 0 ε2 = ∆ε
and
∑n
e−εn/τ = nQV
we can see from Equation 1 that the chemical potential is given as
µ= τ ln
(
nnQZint
)
= τ[
ln
(
nnQ
)
− lnZint
]
which simply becomes
µ= τ[
ln(n/nQ)− ln(1+e−∆ε/τ]
b) The free energy is defined as
Fint = −Nτ lnZint = −Nτ ln(1+e−∆ε/τ)
and the total free energy is given by
Ftot = Nτ[
ln
(
nnQ
)
− ln(1+e−∆ε/τ)−1
]
c) The Entropy is defined as
σint =
(
∂F∂τ
)
V,N= N
∂∂τ
[
τ ln(1+e−∆ε/τ)]
= N
[
ln(1+e−∆ε/τ)+ τ∂∂τ
[ln(1+e−∆ε/τ]
]
where∂∂τ
[ln(1+e−∆ε/τ)] =∆ετ2
e−∆ε/τ
1+e−∆ε/τ
thus we find the entropy to be
σint = N
[
ln(1+e−∆ε/τ)+∆ετ
(
e−∆ε/τ
1+e−∆ε/τ
)]
and the total entropy is defined as
σtot = N
[
ln(nQ
n
)
+52
+ ln(1+e−∆ε/τ)+∆ετ
(
e−∆ε/τ
1+e−∆ε/τ
)]
126
d) The pressure is given as
p = −(
∂Ftot
∂V
)
we can see fromb that the total free energy is given as
Ftot = Nτ[ln(n)− ln(nQ)− ln(1+e−∆ε/τ)−1] = Nτ[
ln
(
NV
)
− ln(nQ)− ln(1+e−∆ε/τ)−1
]
we can see that the derivative in terms of the volume removes every term except the first one, thus
∂F∂V
= −NτV
thus the pressure is given as
p =NτV
e) The heat capacity at constant pressure is given as
CP =52
N+ τ(
∂σint
∂τ
)
P
where the first term comes from the ideal gas solution withoutany degrees of freedom. Thus weneed to solve
∂σint
∂τ= N
∂∂τ
[
ln(1+e−∆ε/τ)+∆ετ
(
e−∆ε/τ
1+e−∆ε/τ
)]
which becomes
∂σint
∂τ= N
[
∆ετ2
e−∆ε/τ
1+e−∆ε/τ −∆ετ2
(
e−∆ε/τ
1+e−∆ε/τ
)
+∆ετ
∂∂τ
(
e−∆ε/τ
1+e−∆ε/τ
)]
= N∆ετ
∂∂τ
(
e−∆ε/τ
1+e−∆ε/τ
)
= N∆ετ
[
(1+e−∆ε/τ)∆ετ2 e−∆ε/τ −e−∆ε/τ ∆ε
τ2 e−∆ε/τ
(1+e−∆ε/τ)2
]
= N(∆ε)2
τ3
[
e−∆ε/τ
(1+e−∆ε/τ)2
]
thus we find that the heat capacity at constant pressure to be given as
CP = N
[
52
+(∆ε)2
τ2
(
e−∆ε/τ
(1+e−∆ε/τ)2
)]
Problem # 6 Isentropic relations of ideal gas
127
a) Show that the differential changes for an ideal gas in an isentropic process satisfy
dpp
+ γdVV
= 0dττ
+(γ−1)dVV
= 0dpp
+γ
1− γdττ
= 0
whereγ = CP/CV . These relations apply even if the molecules have internal degrees of freedom.
We know that for an isentropic gas
pVγ = constant τγ/(1−γ)p = constant τVγ−1 = constant
thus we can see that∂
∂V(pVγ) = Vγ dp
dV+ pγVγ−1 = 0
dpdV
= −γpV
thus we can see thadpp
+ γdVV
= 0
for the temperature and the volume relationship we can use
∂∂V
(τVγ−1) = Vγ−1 dτdV
+ τ(γ−1)Vγ−2 = 0
thusdττ
+(γ−1)dVV
= 0
and finally for the temperature pressure relationship we canuse
∂∂τ
(τγ/(1−γ)p) = τγ/1−γ dpdτ
+ pγ
1− γτ( γ
1−γ−1)= 0
thus we can see that this simplifies intodpp
+γ
1− γdττ
b) The isentropic and isothermal bulk moduli are defined as
Bσ = −V
(
∂p∂V
)
σBτ = −V
(
∂p∂V
)
τ
Show that for an ideal gasBσ = γp; Bτ = p. The velocity of sound in a gas is given asc= (Bσ/ρ)1/2;there is very little heat transfer in a sound wave. For an ideal gas of molecules of massM we havep = ρτ/M, so thatc = (γτ/M)1/2. Hereρ is the mass density.
We have shown thatdpdV
= −γpV
thus the isentropic bulk moduli is given as
Bσ = −V
(
∂p∂V
)
σ= γp
128
for the isothermal bulk moduli we must consider
pV = Nτ
and we know that for a constant temperature
dpdV
V + p = 0dpdV
= − pV
thus the bulk moduli is given as
Bτ = −V
(
∂p∂V
)
τ= p
Problem # 7 Convective isentropic equilibrium of the atmosphere
The lower 10-15 km of the atmosphere-the troposphere-is often in a convective steady state at constantentropy, not constant temperature. In such equilibriumpVγ is independent of altitude, whereγ = CP/CV .Use the condition of mechanical equilibrium in a uniform gravitational field to
a) Show thatdT/dz= constant, wherez is the altitude. This quantity, important in meteorology, is calledthe dry adiabatic lapse rate. (Do not use the barometric pressure relation that was derived in Chapter5 for an isothermal atmosphere.)
We know that the condition of mechanical equilibrium in a uniform gravitational field is given as
Vdp= −Nmgdz
where the left hand term is for the energy due to the expansionand the right hand term is due the thegravitational potential. We can see that
dpdz
= −NmgV
and we also know thatdTdz
=dTdp
dpdz
wheredτdp
= −1− γγ
τp
dTdp
= −1− γγ
Tp
thusdTdz
=1− γ
γTp
NmgV
but we know that
V =NkBT
p
thus we find thatdTdz
=1− γ
γmgkB
which is constant.
129
b) EstimatedT/dz, in 0C per km. Takeγ = 7/5.
Using our previous result we find, also usingγ = 7/5 yields
dTdz
= −27
mgkB
we find the molecular mass of air to be 28.97 amu, thus we find
dTdz
= −27
28.97×1.66×10−27kg(9.8ms−2)
1.38×10−23(m2kgs2K−1)= −9.75
Kkm
= −9.75C
km
c) Show thatp ∝ ργ, whereρ is the mass density.
For an isentropic process we know thatpVγ = constant
we know that the mass density is defined as
ρ =MV
= mNV
thus the volume is given as
V =mNρ
which allows us to substitute into the previous expression as
p
(
mNρ
)γ= constant
thus we can see that
p =ργ
(mN)γ
sop ∝ ργ
If the actual temperature gradient is greater than the isentropic gradient, the atmosphere may be unsta-ble with respect to convection.
Problem # 8 Ideal gas calculations
Consider one mole of an ideal monotomic gas at 300 K and 1 atm. First, let the gas expand isothermallyand reversibly to twice the initial volume; second, let thisbe followed by an isentropic expansion fromtwice to four times the initial volume.
• Process 1 is isothermal and reversible thus we know that the pressure after expansion is given as
piVi = pfVf pf =12
pi
thus the final pressure is half of the initial pressure. We also know that the entropy is given as
∆σ = σ2−σ1 = N ln(V2/V2)
130
the work done by the gas in the expansion is given as
W = −Z V2
V1
pdV =Z V2
V1
(Nτ/V)dV = −Nτ ln(V2/V1)
and the heat is found by assuming that
Q+W = 0 Q = −W = Nτ ln(V2/V1) (6.4)
• Process 2 is an isentropic process, which is a process that happens under constant entropy. we knowthat under isentropic conditions that
∆σ = 0 ∆Q = 0
we can find the temperature using the entropy
σ(τ,V) = N(lnτ3/2 + lnV +constant)
so that the entropy remains constant if
lnτ3/2V = constant
thusτ3/2V = constant
so for an expansion at constant entropy fromV1 toV2 we have
τ3/21 V1 = τ3/2
2 V2 (6.5)
a) How much heat (in joules) is added to the gas in each of the two processes?
From Equation 2 we can see that the heat added for process 1 is given as
Q1 = Nτ ln(2) = NkBT ln(2)
where we know that this can also be expressed as
Q1 = nRTln(2)
whereR is the molar gas constant, thus we find
Q1 = 1 mole×8.31 J/K mole×300 K× ln2≈ 1728 Joules
and for process 2 we have shown that the heat added is
Q2 = 0
b) What is the temperature at the end of the second process?
131
The final temperature for process two can be found by using Equation 3, as
τ f = τi
(
V1
V2
)2/3
or
Tf = Ti
(
V1
V2
)2/3
= 300 K
(
12
)2/3
thus the final temperature isTf = 189 K
c) Suppose the first process is replaced by an irreversible expansion into a vacuum, to a total volume twicethe initial volume. What is the increase of entropy in the reversible expansion, in joules per kelvin?
If we replace process 1 by an irreversible expansion into a vacuum we know that the following propertieshold
∆U = 0 W = 0 Q = 0
no means of doing external work is provided so that the work done is zero. If no work is done than no heatis added in the expansion. Because the energy is also unchanged than the temperature is unchanged. Theincrease of entropy when the volume goes fromV1 toV2 is given by
σ2−σ1 = N ln(V2/V1) = N ln(2)
where
N = 1 mole6.02×1023 atoms
mole= 6.02×1023 atoms
thus the change in entropy is given as∆σ = 4.17×1023
in standard units this is simply∆σ = kB×4.17×1023
this is simply
∆σ = 5.75JK
132
Chapter 7
Fermi and Bose Gases
Roadmap
a) Fermi Gas
1. Metals: Fermi level, density of states, ground state energy, pressure of a fermi gas, heat capacity,and paramagnetic succeptibility.
2. White Dwarfs
3. Nuclei
b) Bose Gas
1. Bose Einstein condensation in liquid helium 4
2. Chemical potential and the temperature at which this condensation occurs
7.1 Fermi Gas
we know that the quantum concentration is given by
nQ =( mτ
2π~2
)3/2
we know that whenn > nQ then this is when quantum behaviours become important. We can also writethis as mτ
2π~2 < n2/3
or
τ <2π~
2
mn2/3 = τ0
if τ ≪ τ0 we call this a degenerate gas and whenτ ≫ τ0 we call this a classical gas.
133
7.1.1 Metals
Metals contain “free” electrons that are donated by atoms inthe lattice. This gives properties of higheloectric conductivity. Lets consider two kinds of metals.The alkili metas which are Li, Na,K and Caeach donate 1 s electrons where for N atoms we have N free electrons. The alkili earth Be,Mg,Ca,Sr, andBa each donate 2 s electrons. Thus N atoms give 2N free electrons. At room temperature, the mean freepath is≈ 40 nm or > 100 lattice spacings. At low temperatures, the meanfree path≈ 400,000 nm = 400µm, thus we can see that the electrons are very free. We will nowderive theFermi energy
7.1.2 Fermi Energy
Lets take a box that have a set of well defined energy levels. Weknow that since it is fermions that we areconstrained by the exclusion principle. The lowest energy level is called the fermi energy levelεF . Weknow that the energy in the box is given by
ε =~
2π2
2mL2(n2x +n2
y +n2z) =
~2π2
2mL2n2
to relate the number of states in energy to the number of states that lie inn andn+dn
N(n)dn= 2× 18
4πn2dn= πn2dn
thus we can see that
εF =~
2
2mL2π2n2F
7.1.3 Fermi Gas
the Fermi energy is given as
εF =~
2
2m
(
3π2NV
)3/2
the density of states is given as
D(ε) =V2π
(
2m~2
)3/2
ε1/2
the Fermi density of states is given as
D(εF) =32
NεF
and the energy is given as
u(0) =35
NεF =35
NkBT
7.1.4 Pressure of the Fermi gas (T = 0)
We recall that the pressurep is given as
p = −(
∂u(0)
∂V
)
σ,N
134
once again we haveu(0) can be written as
u(0) =32
NεF =35
[
~2
2m(3π2)2/3
]
N5/3V−2/3
so we simply need to take the derivative to find
p = −35
(
−23
)[
~2
2m(3π2)2/3
](
NV
)5/3
=25
[
~2
2m
(
3π2NV
)2/3]
NV
where
εF =
[
~2
2m
(
3π2NV
)2/3]
thus the pressure is given as
p =25
εFNV
or
pV =25
NkBTF
it might be instructive to see how big this pressure really is. Lets work out the pressure for an ideal gas
7.1.4.1 Pressure of ideal gas at STP
we know that the pressure is given as
p =RTV
=8.3 J mole−1K−1×233
22.4×10−3m3 ≈ 105Pa
now what how does this relate to the Fermi pressure? Lets takesodium as our example, thus
V = 23 cm3 = 23×10−6m3 TF = 35,000 K
thus
p =25
RTF
V≈ 5×109 Pa
thus the pressure of the Fermi gas to the pressure of an ideal gas at STP is given as
pF
p≈ TF
T× 22,400
23≈ 105
7.1.5 Specific Heat of Fermi Gas
For insulators at room temperature isCV = 3NkB and for metals the result is basically the same.
135
7.1.5.1 Approximate argument forCV
the number of lectrons that are excited is given by an increase in temperature is
∼ NkBTεF
∼ NTTF
thus the change in the energy is
∼ NTTF
kBT
and thus the heat capacity is
CV =∂
∂T
(
NkBT2
TF
)
∼ NkBTTF
∼ TTF
×classical value∼ 0.01(NkB)
for sodium at room temperature. The exact result is given as
CV =π2
2NkBT
TF= γT
whereγ is given as
γ =π2
2NkB
TF
the total heat capacity of a metalCV = γT +AT3
where the first term come from the electrons and the second term comes from the phonon contribution. Atroom temperature, the phonos dominate and at low temperature, the elctrons dominate.
7.1.6 Paramagnetic Susceptibility
Calssically
χ =Nµ2
kBTCurie’s Law
when we increaseeB, only electrons within∼ kBT of εF ca flip spins, thus
χF ∼ Nµ2B
kBT
(
TTF
)
=Nµ2
B
kBTF
this is known as Pauli paramagnetism.
7.1.7 White Dwarfs
White dwarfs are stars that have undergone gravitational collapse and is help up under degeneracy pressure.The density is typically 105−106gm cm−3. This was first discovered from Sirius B which was empericallydetermined from the pertubations from Sirius B. We find that the mass of Sirius B isMB = 2×1033g andthe radius is given asrB = 2×109cm.This can be acquired from the Planck distribution and Wiens’s law.From this, we find that the densityρ ∼ 7×104g cm−3. We believe that all matter is ionized and thus we
136
have Fermi gases. We must consider the protons and electronsas being two distinct samples of Fermigases. If we have
ρ ∼ 1 g cm−3 l ∼ 10−10 m
ρ ∼ 106 g cm−3 l ∼ 10−12m
andNV
∼(
110−12
)3
m−3 ∼ 1036m−3
and the Fermi energy is
εF ∼ εF
(
1036
1028
)2/3
≈ 3×105eV ⇒ TF ≈ 3×109K
where we know that the temperature needed for fusion isT ∼ 107 K. SinceT ≪ TF this is a highlydegenerate gas. Lets do this for protons, lets first assume that the nuclie are protons. The mass of theproton is approximately 2000me. We can see that
εF ∝1m
thus for the protons
TF ∼ 3×109
2000∼ 1.5×106K
which is treated as a classical gas.
7.1.8 Neutron Stars (The Nucleus)
The atomic number A is the number of protons and neutrons. Theradius is given as
r = (1.3×10−13cm)A1/3
thenNV
=3A
4π(1.3×10−13cm)3A≈ 1038cm−3 ∼ 1044m−3
sinceNp ≈ Nn thusNV
∼ 5×1043m−3
plugging in the numbers yield
εF = 3×107eV∼ 30 MeV TF ≈ 4×1011 K
7.1.9 Bose-Einstein Condensation in Liquid4He
The helium nucleus has 2 neutrons and 2 protons with a spin of 0and obeys Bose-Einstein statistics.This was historically discovered when people started measuring the heat capacity. They measured thatsomething spectacular happened as you lowered the temperature of the helium liquid, when theyt got to2.18 K they saw a weird behavior in the heat capacity. They named this transition point, the lambda point.
137
Justification for treating Helium 4 as a gas
We will treat Helium 4 as a gas rather than a liquid in order to model its charactersitics. there are threemain points
• Computer simulations predict the density from known quantum properties. In fact, the measureddensity is about three times less. Interactions will be weakened as they are pushed apart. Thezero point fluctuations in the atoms increase their kinetic energy, thus pushing them further apart(increase their seperation).ρ ≈ 0.1 g/cm−3.
• Even forT > Tλ, the thermal conductivity and the viscosity are roughly thesame as those of a gasof the same density.
• It is very hard to solidify helium, helium liquid will alwaysremain liquid at atmospheric pressuredown to arbitrarily low temperatures (lowest temperaturespossible). This implies weak interactionsbetween helium atoms. To solifify helium you must apply a pressure greater than 25 atm.
Lets consider 1 cm3 cube, the ground state energy is given as
ε0 =π2
~2
2mL2(12+12+12) =3π2
~2
2mL2
the first excited state is given as
ε1 =6π2
~2
2mL2
and the difference is
ε1− ε0 =3π2
~2
2mL2 ≈ 10−37J≈ 10−18eV≈ 10−14K
if we were to cool this to 10−16 K (impossible) vertually all of the atoms will be i the groundstate. In fact,at 1 K virtually all of the atoms are in the ground state. The entire behavior of this gas is controlled mainlyby the chemical potential. The above calculation is wrong because it is a Bose gas.
7.1.10 Bose Einstein Distribution
The BE distribution si given as
f (ε) =1
e(ε−µ)/τ −1
it is a requirment that the chemical potential always be lessthen the energy (µ < ε). For convinience wewill set
ε0 = 0 f (0,T) =1
e−µ/kBT −1lets suppose that the temperatureT becomes very small. In the limit
T → 0 f (0,0) = N
whereN is the number of atoms in the box, so
f (0,0) = N = limT→0
1
e−µ/τ −1where µ/kBT ≪ 1
138
we are really concerned with the behavior of the chemical potential rather than the temperature. Given theabove condition allows us to expand the exponetial
11− (µ/kBT)−1
≈−kBTµ
hence
µ≈−kBTN
this tells us that the chemical potential is just below the ground state energy, but very very close to it.Henceµ< 0, but very close toε0 = 0.
Example
lets takeT = 1 K N ∼ 1022
thusµ≈−10−45J
which is very close to the ground state energy. Thus the probability of the number of particles in the firstexcited state is as follows
f (ε1,1K) =1
e(ε!−µ)/τ −1≈ kBT
ε1−µ≈ 1 K
10−14 K≈ 1014
thus, at 1 K only 1 atom in 108 is in the first excited state. We can finally calculate the occupancy of theground state as a function of temperature.
7.1.11 Temperature dependence off (0,T)
The first thing we need to do is write down the density of states, this is given as
D(ε) =V
4π2
(
2m~2
)3/2
ε1/2 spin 0
once again lets consider a few energy levels, a ground state and higher energy states, where the totalnumber of particles is given as
N = N0+Ne
whereN0 ar the atoms in the ground state andNe are the atoms in the excited state. Thus
N = N0(T)+Z ∞
0D(ε) f (ε,T)dε
we must now evaluate this, this is equal to
N =1
e−µ/kBT −1+
V4π2
(
2m~2
)3/2 Z ∞
0
ε1/2
e(ε−µ)/kBT −1dε
as before, we can neglect the chemical potential because it is negligable, thus we have
x =ε
kBT
139
which allows us to write
Ne(T) =V
4π2
(
2m~2
)3/2
(kBT)3/2Z ∞
0
x1/2
ex−1dx=
V4π2
(
2m~2
)3/2
(kBT)3/2(1.31π1/2)
which can be simplified as
Ne = 0.17V
(
mkBT~2
)3/2
the fraction of excited states is
Ne
N≈ 0.17
VN
(
mkBT~2
)3/2
≈ 2.61nQ
n
this a very useful simple resultNe
N≈ 2.61
nQ
n
wha this shows us is that the number of excited atoms increases as the temperature to the 3/2. This givesus a prediction of the transition point for helium.Tλ is given by
Ne
N= 1
thus
Tλ =~
2
2m
(
N0.17V
)2/3
≈ 3 K
where the observed value isTλ = 2.18 K. If we write
Ne
N=
(
TTλ
)3/2
andN0
N= 1−
(
TTλ
)3/2
qualitatively this result is correct, but quantitatively it is incorrect. The model is too simple. What wehave seen is that at low temperatures we have a high concentration of atoms in the ground state. Thus wehave what is called a macroscopic quantum state. The Bose-Einstein condensation is this, that is to say,we have some (one) quantum mechanical wave function
Ψ(~r, t) = |ψ(r, t)|eiφ(r,t)
for all atoms in the ground state. There are two other kinds ofsystems that have the same properties.Helium 3 atoms are fermions but it turns out that at low enoughtemperatures these helium 3 atoms canpair together to give bosons and they can also undergo a condensation to the ground state to create asuper fluid (1972). The third example is super conductivity,this is where we have electrons that becomecooper pairs which act like bosons. (~k ↑,−~k ↓), where they have zero momentum and zero spin. This wasunderstood theoretically in 1957 BCS.
140
7.1.12 Properties of4He Superfluids: Thermomechanical effects
Lets suppose we have two resevoirs of helium that are connected by a narrow capillary (pipe). It blocksnromal atoms but it passes “superfluid” atoms. We then cool the helium down and put pressure on oneside of the resevoir. We see that we will have an abundance of superfluid atoms on one side of the resevoir.The side which contains the most superfluid atoms will lower its temperature while the side that hasless superfluid atoms will increase its temperature. Suppose we take the same equipment and increase thetemperature on one side. Initially both resevoirs are at temperatureT and we then increase the temperatureby ∆T on one side. This will create an osmotic pressure of superfluid helium. The level on one side fallswhile the level on the other side increases to take care of thepressure difference.
7.2 Summary
For the Fermi gas
εF =~
2
2m
(
3π2NV
)2/3
the density of states is given as
D(ε) =V
2π2
(
2m~2
)3/2
ε1/2
and
D(εF) =3N2εF
the ground state energy is given as
u(0) =35
NεF
also
pV =25
NkBTF
the heat capacity is given as
CV =π2
2NkBT
TF= γT
and also
χ ∼ µ2B
kBTF
for the Bose gas
T → 0 µ= −kBTN
andNe
N≈ 0.17
VN
(
mkBT~2
)3/2
≈ 2.61nQ
n
141
7.3 Problems
Problem # 1 Energy of relativistic Fermi gas
For electrons with an energyε ≫ mc2, wherem is the rest mass of the electron, the energy is given byε ≃ pc, wherep is the momentum. For electrons in a cube of volumeV = L3 the momentum is of theform (π~/L), multiplied by(n2
x +n2y +n2
z)1/2, exactly as for the nonrelativistic limit.
a) Show that in this extreme relativistic limit the Fermi energy of a gas ofN electrons is given by
εF = ~πc(3n/π)1/3 = ~πc
(
3NVπ
)1/3
wheren = N/V.
Since we know that we are in the relativistic case that the energy is defined as
ε =π~cL
n
but we know that the Fermi density is related to the number of particles by
N = 2× 18× 4
3πn3
F =π3
n3F nF =
(
3Nπ
)1/3
The factor of two arises because an electron has two possiblespin orientations. The factor of 1/8 arisesbecause only triplets in the positive ocatant of the sphere in n space are to be counted, thus the Fermienergy is given by
εF =~πcL
nF = ~πc
(
3NL3π
)1/3
which simplifies to
εF = ~πc
(
3NVπ
)1/3
= ~πc(3n/π)1/3
b) Show that the total energy of the ground state of the gas is
U0 =34
NεF
We know that this can be solved by knowing that the total energy of the system in the ground state is givenby
U0 = 2 ∑n≤nF
εn = 2× 18×4π
Z nF
0n2dnεn =
π2~cL
Z nF
0n3dn=
π2~c
4Ln4
F =π2
~c4L
n3FnF =
3π~c4L
NnF
which simplifies into
U0 =34
NεF
142
Problem # 2 Energy, heat capacity, and entropy of degenerateboson gas
Find expressions as a function of temperature in the regionτ < τE for the energy, heat capacity, andentropy of a gas ofN noninteracting bosons of spin zero confined to a volumeV. Put the defenite integralin dimensionless form; it need not be evaluated. The calculated heat capacity above and belowτE isshown in Figure 7.12. The difference between the two curves is marked: It is ascribed to the effect ofinteractions between the atoms.
We know that the density of modes is given by
D(ε) =V
4π2
(
2m~2
)3/2
ε1/2
for a particle of spin 0. The total number of atoms in the ground and excited orbitals is given by the sumof the occupancies of all orbitals
N = ∑n
fn = N0(τ)+Ne(τ)
and the total energy is given byU = ∑
nεn fn = N0ε0(τ)+Neε(τ)
but we know that the number of atoms in the excited state is given by
Ne(τ) =Z ∞
0D(ε) f (ε,τ)dε
thus the total energy is given by
U =Z ∞
0D(ε) f (ε,τ)ε(τ)dε
we also know that the Bose-Einstein distribution is given as
f (ε,τ) =1
e(ε−µ)/τ −1
thus we find that the total energy is given as
U =V
4π2
(
2m~2
)3/2 Z ∞
0
1
λ−1eε/τ −1ε3/2dε
whereλ = eµ/τand if we letx = ε/τ we find
U =V
4π2
(
2m~2
)3/2
τ5/2Z ∞
0
1λ−1ex−1
x3/2dx
For λ we can see thatλ = 1
this is because we can Taylor expand the activity to find
λ ≈ 1+µτ
= 1− 1N
143
but we have seen that−µ=
τN
N ≫ 1
The integral can now be written as
Z ∞
0
e−x
1−e−xx3/2dx=∞
∑s=1
Z ∞
0x3/2e−sxdx
if we let y = sxthen∞
∑s=1
Z ∞
0x3/2e−sxdx=
(
∞
∑s=1
s−5/2
)
Z ∞
0e−yy3/2dy
if we now lety = u2 then we findZ ∞
0e−yy3/2dy= 2
Z ∞
0e−u2
u4du=34
√π
and we also find that(
∞
∑s=1
s−5/2
)
≈ 1.34
thus we find that the total energy is given by
U =4.02V
16
(
2mπ~2
)3/2
τ5/2
and we know that the heat capacity is given by
CV =∂U∂τ
=52
4.02V16
(
2mπ~2
)3/2
τ3/2
which simplifies into
CV = 0.63V
(
2mπ~2
)3/2
τ3/2
and the entropy is given by
CV = τ(
∂σ∂τ
)
V
thus we can see that the total entropy is given by
σ =Z τ
0
CV
τdτ = 0.61V
(
2mπ~2
)3/2 Z
τ1/2dτ =23(0.63)V
(
2mπ~2
)3/2
τ3/2
or
σ = 0.42V
(
2mπ~2
)3/2
τ3/2
Problem # 3 Fluctuations in a Fermi gas
144
Show for a single orbital of a fermion system that
〈(∆N)2〉 = 〈N〉(1−〈N〉)
if 〈N〉 is the average number of fermions in that orbital. Notice that the fluctuation vanishes for orbitalswith energies deep enough below the Fermi energy so that〈N〉 = 1. By definition,∆N = N−〈N〉
From Equation 5.83 we know that the fluctuations go as
〈(∆N)2〉 = τ∂〈N〉
∂µ
Equation 6.2 gives that the average number of particles is given by
〈N(ε)〉 = f (ε) =1
e(ε−µ)/τ ±1
where the + is for fermions and the - is for bosons,f (ε) is the average occupancy that denotes the thermalaverage number of particles in an orbital of energyε. Thus for a Fermi gas we find that
〈N〉 =1
e(ε−µ)/τ +1
and∂〈N〉
∂µ=
ddµ
(
1
e(ε−µ)/τ +1
)
=ddµ
(
1K
)
if we let
K = e(ε−µ)/τ +1dKdµ
= −1τ
e(ε−µ)/τ
thusdNdµ
=dNdK
dKdµ
and since we know that
N =1K
dNdK
= − 1K2
we find∂N∂µ
=1τ
e(ε−µ)/τ
(e(ε−µ)/τ +1)2
thus the fluctuations is given as
〈(∆N)2〉 =e(ε−µ)/τ
(e(ε−µ)/τ +1)2
to write it in a more suggestive form we can simply do
〈(∆N)2〉 =e(ε−µ)/τ +1−1
(e(ε−µ)/τ +1)2=
1
e(ε−µ)/τ +1
(
1− 1
e(ε−µ)/τ +1
)
which can also be written as〈(∆N)2〉 = 〈N〉(1−〈N〉)
Problem # 4 Fluctuations in a Bose gas
145
If 〈N〉 as in (11) is the average occupancy of a single orbital of a boson system, then from 5.83 show that
〈(∆N)2〉 = 〈N〉(1+ 〈N〉)
thus, if the occupancy is large, with〈N〉 ≫ 1, the fractional fluctuations are of the order unity:〈(∆N)2〉/〈N〉 ≈ 1, so that the actual fluctuations can be enormous. It has beensaid that “bosons travel inflocks.”
As before we know that
〈N〉 = f (ε) =1
e(ε−µ)/τ −1
and as before∂N∂µ
=1τ
e(ε−µ)/τ
(e(ε−µ)/τ −1)2
thus the fluctuation is given as
〈(∆N)2〉 =e(ε−µ)/τ
(e(ε−µ)/τ −1)2
written in a more suggestive way
〈(∆N)2〉 =e(ε−µ)/τ −1+1
(e(ε−µ)/τ −1)2=
1
e(ε−µ)/τ −1
(
1+1
e(ε−µ)/τ −1
)
which can also be expressed as
〈(∆N)2〉 = 〈N〉(1+ 〈N〉)
146
Chapter 8
Heat and Work
Roadmap
1. Heat and Work: Exact and inexact differentials
2. Heat engines and refrigerators: 2nd law revisited
3. Carnot cycle for the ideal gas
4. Work and heat at constant temperature or at constant temperature and pressure: Enthalpy and Gibbsfree energy
8.1 Heat and Work: Exact and inexact differentials
We know thatheat is a transfer of energy by thermal contact, which is usually denoted asdQ. Work isa transfer of energy by changes of external parameters, i.e volume, electric field or magnetic fields. Forreversible process, the first law tell us that the total change is given by
du= dQ+dW
where the first term is for the heat and the second term is for the work. du is what we call an exactdifferential. If we start at some pointa to some pointb it doesnt matter what path we take, the change inthe energy is independent of the path. But for the change in heat or the cahnge in the work are dependent onthe path chosen. The internal energyu is a function of state while the heat and the work are not functionsof states. An exact differential is of the following form
dF = F(x+dx,y+dy)−F(x,y) = A(x,y)dx+B(x,y)dy
anything that can be written in this form is an exact differential. But not all functions of the formC(x,y)dx+ D(x,y)dy are of the form as beforeG(x+ dx,y+ dy)−G(x,y). We want to now take twoexamples to illustrate this point.
8.1.1 Example 1
Lets letdG= αdx+β
xydy= αdx+βx(d lny)
147
whereα andβ are cosntants. If we now take the integral, via (1,1) to (1,2)to (2,2)Z f
idG= α+2β(ln2− ln1) = α+2β ln2
If we now take the integral, via (1,1) to (2,1) to (2,2)Z f
idG= β(ln2)+α = α+β ln2
this is an example of an inexact differential. The real problem is in the term ofβ. Now for an exactdifferential
dF =dGx
=αx
dx+βdyy
= d(α lnx+β lny)
now if we dow the same thing as last time we find If we now take theintegral, via (1,1) to (1,2) to (2,2)Z f
idG= α ln2+β ln2 = α ln2+β ln2
If we now take the integral, via (1,1) to (2,1) to (2,2)Z f
idG= α ln2+β ln2 = α ln2+β ln2
where we can see that these are exactly the same. Any type of work can be converted to any other type ofwork (in principle) and entropy is conserved. Lets imagine that we are compressing a gas in a cylindervia some magnetic interaction. If we attach a bar magnet to the piston and then we move a secondmagnet to the first which compresses the gas, thus we have usedmagnetic work to create mechanicalwork. Work can be also be completely converted into heat, i.estirring your coffee in the morning.However, the converse is not true, heat cannot be converted completely into work in a reversible process.Thus it is really the study of this third point that is the subject of study of heat engines and refrigerators.
8.2 Heat Engines and Refrigerators
8.2.1 Engines
This is some kind of cyclic device which takes in heat to convert to mechanical work and reject some heatat a lower temperature. We have some thermal resevoir at sometemperatureT1 with some other resevoir atT2, and we also have heatQ1 andQ2 and we get out the workW. WhereQ1,Q2, andW are all magnitudes.The entropy is given as
σ1 =Q1
τ1σ2 =
Q2
τ2
hence, the workW is simply
W = Q1−Q2 = Q1−τ2
τ1Q1 = Q
(
τ1− τ2
τ1
)
where this is known as the Carnot efficiency
ηC =WQ
=τ1− τ2
τ1
this is the maximum possible efficiency that an engine can runon. In practice all engines will haveadditional losses and this efficiency can never be achieved.A car engine whereτ1 = 600 K andτ2 = 300K, we can see that the maximum efficiency is about 1/2. We can reverse the process to make a refrigerator.
148
8.2.2 Refrigerators
A refrigerator is simply a heat engine in reverse. So we have aresevoir atT1 and another atT2, we alsohaveQ1 andQ2 and we put in workW. The Carnot efficiency is the following,
γC =Q2
W=
Q2
Q1−Q2
the exact relationships for the entropy still hold, hence
γC =Q2
τ1τ2
Q2−Q2=
τ2
τ1− τ2
this is the maximum efficiency of a refrigerator. Lets suppose thatτ1 = 300 K andτ2 = 260 K, we can seethat
γC =26040
= 6.5
This is also how air conditioners work, whereT2 is the temperature in the room andT1 is the temperatureof the outside. We can also consider aheat pump whereT2 is the temperature outside andT1 is thetemperatureT2 and we find that the Carnot efficiency is now given by
γC =Q1
W=
Q1
Q1− τ2τ1
Q1=
τ1
τ1− τ2
we can take an example ofτ2 = 270 K andτ1 = 300 K, thus the Carnot efficiency isγC = 10. We can seethat a heat pump is very efficient.
8.2.3 Non ideal case for Heat engine
Just to reacp very briefly, we know that
∆σ = −Q1
τ1+
Q2
τ2= 0 ⇒ Q2 = Q1
τ2
τ2
and the efficiency is given as
ηC =WQ1
=τ1− τ2
τ1
if we have losses then∆σ > 0 σ2 > σ1
if we now go back we can see that
Q2 > Q1τ2
τ1
for eaxmple, lets set
Q2 = Q1
(
τ2
τ1+δ)
whereδ is the inefficiency, thus we can see that
η =Q1−Q1
(
τ2τ1
+δ)
Q1< ηC
149
thus we can see that the Carnot efficiency is the maximum one can get from heat engines.
η ≤ ηC γ ≤ ηC
these are statements of the second law of thermodynamics. Thus
∆σ ≥ 0
8.3 Carnot Cycle
8.3.1 The Ideal Gas
If we assume that we have an isothermal expansion of a gas fromσlow to σhi and a fixed temperatureτhi. We then do a second expansion which is adiabatic fromσhi andτhi to τlow. We then have to do twodifferent compressions, the first one is isothermal going from σhi to σlow and atτlow, we then do anotheradiabatic compression fromτlow to τh at σlow. This are the four steps of the Cranot cycle. We know thataround the entire cycle, the change in energy is equal to zero
0 =I
du
this must also be equal toI
du=
I
τdσ−I
pdV = 0
where the first term is for the heat and the second term is for the work. Thus the work done by the gas is
W =I
τdσ = τhi(σlow−σhi)+ τlow(σhi −σlow)
and thus the work done by the gas is given as
Wgas= (τhi− τlow)(σlow−σhi)
which is simpy the area of a rectangle. We can replicate this entire process forN molecules and gas usinga p andV diagram. The firs thing is to expand the gas isothermally, thus we know that the pressure dropsand the volume increases. In order to do this we have to apply heatQ1. We then do an adiabatic expansionwhere the volume increases more and the pressure drops more.We then do an isothermal compression atthe low temperature end, thus the volume decreases and the pressure increases, and thus heat is rejected.Finally we do the adiabatic compression to get back to the starting point. As before, the area tells us thework done by the gas. We want to figure out thework done by the gasin each of these 4 processes
W1→2 =Z V2
V1
pdV = Nτhi
Z V2
V1
dVV
= Nτhi ln
(
V2
V1
)
= Q1
then we have the work done during the adiabatic expansion
W2→3 = [u(τhi)−u(τlow)] =32
N(τhi − τlow)
next we go from point 3 to point 4, which is an isothermal expansion
W3→4 = Nτlow ln
(
V4
V3
)
= Q2
150
and finally we find
W4→1 = [u(τlow)−u(τhi)] =32
N(τlow− τhi)
and the total work is given as
W = Nτhi ln
(
V2
V1
)
+Nτlow ln
(
V4
V3
)
The goal of this calculation is to rederive the Carnot efficiency for this process. To find the Carnotefficiency we must find a relationship amongV1...V4. Lets assume that we have an ideal monotomic gas.We then want to use a relationship derived a while ago that we said was useful. At constant entropy wefound the following result
τV2/3 = constant
we now consider where we have processes at constant entropy,this is for 2 going to 3 and we start with
τHV2/32 = τLV2/3
3
in other wordsV3
V2=
(
τH
τL
)3/2
and now if we look at 4 to 1 we findτLV2/3
4 = τHV2/31
thusV4
V1=
(
τH
τL
)3/2
=V3
V2
if we now go back to the expression for the work done we find
V2
V1=
V3
V4
and finally, what we see is the following
ln
(
V2
V1
)
= − ln
(
V4
V3
)
therefore thework done by the gasis given by
W = N(τH − τL) ln
(
V2
V1
)
we finally find the Carnot efficiency
ηC =WQ1
=N(τH − τL) ln(V2/V1)
NτH ln(V2/V1)
which reduces to what we found previously
ηC =τH − τL
τH
151
8.4 Heat and Work at Constant Temperature or Pressure
8.4.1 Isothermal
Isothermal conversion of work into heat. We have a resevoir at a temperatureτ and a piston that willcompress the gas isothermally. For an isolated system, the internal energy is given as
u = u(σ,V)
anddu(σ,V) = τdσ− pdV thermodynamic identity
which reduces todu= d(στ)− pdV
and the free energy isdF(τ,u) = d(u− τσ) = −pdV
which is the change in work for the system. Thus the work done on the gas is
dW = dF
we can see that heat is an inexact differential. These two thermodynamic functions play different roles,wheredF is a function of state whiledW is not. This is why the Helmholtz free energy is such a usefulfunction.
8.4.2 Isobaric (constant pressure)
Boiling water, chemical reactions are usually at constant pressure. If we impoise a constant pressure onour system with a piston at a constant force.
F = pA
initially we have some water and we apply heat from underneath. If we start boiling the water, this willprovide a pressure on the piston from the vapor. This moves the piston by an amountdx which will giveusdV = Ada. We need to supply heat to evaporate the water and expand the volume to accomdidate thegas. We must now write down the work done on the vapor
W = −pdV = −−d(pV) dp= 0
thus the change in the internal energy of the liquid and the vapor is given by
du= dQ−d(pV)
we will now define a new quantityH which is
dH = d(u+ pV) = dQ
where this new quantityH is known as theenthalpy, thus
dQ= dH
152
8.4.3 Work and Heat at constant Pressure
If we take our previous system, where we apply some heat, but now we also apply some work to thesystem, i.e a paddle moving the water. So
du= dQ−d(pV)+dW
we can therefore write this as followsdW = du+d(pV)
and alsodQ= τdσ = d(τσ) constantτ
putting this all together givesdW = d(u+ pV− τσ) = dG
whereG = u+ pV− τσ = F + pV
where this is known as theGibbs Free Energy. Thus the work done on the system is simply
dW = dG
8.4.4 Gibbs Free energy for varyingN
We know thatdG= du+ pdV+Vdp− τdσ−σdτ
recall thatdu= τdσ− pdV+µdN
this is the thermodynamic identity whenN varies. We can now insert this into the previous line to find
dG= −σdτ+Vdp+µdN
so we now have this new term that is very useful. We see immideately that we can write down the followingthree results
−σ =
(
∂G∂τ
)
p,NV =
(
∂G∂p
)
τ,Nµ=
(
∂G∂N
)
τ,p
the last result is also very useful, if we integrate
G(τ, p,N) = Nµ(p,τ)
this shows us that the chemical potentialµ is the Gibbs free energy per particle. This is really what makesphase transformations tick.
153
8.5 Summary
We now summarize the important results. We begin with heat as
dQ= τdσ
is the transfer of energy by thermal contact with a resevoir.We also talked about work, which is anotherinexact differential which is the transfer of energy by the change in external parameters, we also found
dσ = 0
we also found for an ideal heat engine
ηC =τ1− τ2
τ1τ1 > τ2
for a refrigirator we find
ηC =τ2
τ1− τ2can be > 1
we then showed for an isothermal process that the work
dW = dF
we also talked about isobaric heat and work allowed us to introduce theenthalpy H which is given by
H(σ, p) = u+ pV
and the heat of vaporizationdQ= dH
we also introduced the Gibbs free energy
G(τ, p) = u− τσ+ pV = F + pV
and the work done on the system at constant pressure and temperature is
dW = dG
and finally we discussed what happens when the number of particles change, we find
dG(τ, p,N) = −σdτ+Vdp+µdN
and hence, we find three very useful results
−σ =
(
∂G∂τ
)
p,NV =
(
∂G∂p
)
τ,Nµ=
(
∂G∂N
)
τ,p
154
8.6 Problems
Problem # 1 Photon Carnot Engine
Consider a Carnot engine that uses as the working substance aphoton gas. The Carnot cycle can berepresented as
isentropic
1
23
4
hτlτ
lσ
hσ
σ
τ
isot
herm
alisotherm
al
isentropic
The cycle consists of two expansion phases (1→ 2 and 2→ 3) and two compression phases (3→ 4 and4→ 1). There are two isentropic phases and two isothermal phases. The work done is given by the areaof the rectangle created by the solid line and the heat consumed atτh is the area sorrounded by the brokenline. Since we are considering a photon gas as the working substance of the engine we must consider theenergy density of a photon gas which is given as
U =π2V
15~3c3τ4 (8.1)
but we know thatτdσ = dUτ + pdV
thus at constant volume we have
dσ =4π2Vτ2
15~3c3 dτ
integrating both sides yield
σ =445
π2V~3c3τ3 (8.2)
a) Givenτh andτl as well asV1 andV2, determineV3 andV4.
155
We know that the entropy is the same at both 1 and 4 and 2 and 3 thus
σ4 = σ1 σ2 = σ3
using Equation 2 we findV4τ3
l = V1τ3h V2τ3
h = V3τ3l
thus
V4 = V1
(
τh
τl
)3
V3 = V2
(
τh
τl
)3
b) What is the heatQh taken up and the work done by the gas during the first isothermal expansion? Arethey equal to each other, as for the ideal gas?
We know thatQh =
Z
dQ=
Z
pdV =
Z
τdσ
thus
Qh =Z 2
1τdσ = τ(σ2−σ1) =
4π2
45~3c3τ4h(V2−V1) =
445
ατ4h(V2−V1)
where
α =π2
~3c3
so we find that the heat taken up is
Qh =445
ατ4h(V2−V1)
and the work is found by usingdW = dU−dQ
thus the work is
W12 = −(
Z 2
1dU−
Z 2
1τdσ
)
the minus sign comes because this is thework done on the gas, thus we can solve to find
W12 = −(
U2−U1−[
445
ατ4h(V2−V1)
])
the change in energy is given by Equation 1 as
U2−U1 =α15
τ4h(V2−V1)
thus the work is
W12 = −(
α15
τ4h(V2−V1)−
[
445
ατ4h(V2−V1)
])
or
W12 =145
ατ4h(V2−V1)
we can see that these two quantities are not equal as in the case of the ideal gas.
156
c) Do the two isentropic stages cancel each other, as for the ideal gas?
The work for the two isentropic phases is given by
W23 = −Z 3
2dU W41 = −
Z 1
4dU
sincedσ = 0, as before the minus sign comes from the fact that work is being done on the gas we find
W23 = −[U3−U2]
= −(
115
α[τ4l V3− τ4
hV2]
)
W23 = −(
115
α
[
τ4l V2
(
τh
τl
)3
− τ4hV2
])
W23 =115
ατ3hV2[τh− τl ]
and also
W41 = −[U1−U4]
= −(
115
α[τ4l V1− τ4
hV4]
)
W41 = −(
115
α
[
τ4l V1− τ4
hV1
(
τh
τl
)3])
W41 = − 115
ατ3hV1[τh− τl ]
thus
W23+W41 =115
ατ3h[τh− τl ][V2−V1]
we can see that they do not cancel each other out as in the case for the ideal gas.
d) Calculate the total work done by the gas during one cycle. Compare it with the heat taken upτh andshow that the energy conversion efficiency is the Carnot efficiency.
To find the total work done we must finally consider the work done during
W34 = −(
Z 4
3dU−
Z 4
3τdσ
)
= −(U4−U3− τl (σ4−σ3))
which yields
W34 = −(
α15
τ4l (V4−V3)−
445
ατ4l (V4−V3)
)
= −( α
45τ4
l (V3−V4))
= −(
α45
τ4l
(
V2
(
τh
τl
)3
−V1
(
τh
τl
)3))
= − α45
τlτ3h(V2−V1)
157
where the total energy is given by
W = W12+W23+W34+W41
=145
ατ4h(V2−V1)+
115
ατ3h[τh− τl ][V2−V1]−
145
ατlτ3h(V2−V1)
=145
ατ3h[τh− τl ][V2−V1]+
115
ατ3h[τh− τl ][V2−V1]
W =445
ατ3h[τh− τl ][V2−V1]
and comparing this to the Carnot efficiency yields
ηC =WQh
=τh− τl
τh=
Th−Tl
Th
Problem # 2 Geothermal Energy
A very large massM of porous hot rock is to be utilized to generate electricity by injecting water andutilizing the resulting hot steam to drive a turbine. As a result of heat extraction, the temperature of therock drops, according todQ= −MCdTh, whereC is the specific heat of the rock, assume to betemperature independent. If the plant operates at the Carnot limit, calculate the total amountW ofelectrical energy extractable from the rock, if the temperature of the rock was initiallyTh = Ti , and theplant is to be shut down when the temperature has dropped toTh = Tf . Assume that the lower resevoirtemperatureTl stays constant.
At the end of the calculation, give a numerical value, in kWh,for M = 1014 kg (about 30 km3), C =1 J g−1K−1, Ti = 6000C, Tf = 1100C, Tl = 200C. Watch the units and explain all steps! for comparison:The total electricity produced in the world in 1976 was between 1 and 2 times 1014 kWh.
Since we know that this engine operates at the Carnot efficiency then
ηC =Th−Tl
Th=
WQh
thus the work is given as
dW =Th−Tl
ThdQ= −Th−Tl
ThMCdTh
thus the total work is given as
W = −MCZ Tf
Ti
Th−Tl
ThdTh = MC
[
[Ti −Tf ]+Tl ln[Tf
Ti]
]
plugging in numbers given by
Ti = 873 K Tf = 383 K Tl = 293 K
W = 2.48×1016kg Jg
= 2.48×1019J
in kWh this is simply1 kWh= 3.6×106 J
thus the work done is given by
W = 6.88×1012 kWh
158
Chapter 9
Phase Transformations
This is when we transform from one phase to another, .e water to ice, water to vapor. We do this bychanging the temperature and pressure. We can also go from ice to vapor, this is known as sublimation(frost). Another example is going from paramagnetic to ferromagnetic. Also going from normal to super-conducter. Finally normal Helium 4 to superfluid Helium 4.
Road map
1) Gas/liquid/solid phase transformations.
• p vsV isothermal
• p vs T coexistance curves- triple point
2) Clausis-Clapeyon Equation
Latent heat,vp vsT. Change is boiling point with pressure and latent heat of vaporization of ice
3) Van Der Walls Equations
• departures from ideal gas
• predictions ofp vsV
• critical points at which all hree phases can exist
9.1 Gas, Liquid, and Solid Phase Transformations
9.1.1 p vsV
If we look at a plot of the pressure versus the volume for a liquid curve we find that at low volumes wehave an isotherm. The reason is due to the incompressabilityof liquid. We start in a liquid state with apiston pushing hard against it. As we keep removing the pressure we get to a point where the piston israised above the liquid. The pressure than becomes constant, and this is where we can have both liquidand gas (this all occurs inT < Tc). The place where the pressure becomes constant will be denoted p0
and this depends on the temperature, we also get what are called metastable phases. And finally as we
159
decrease the pressure and increase the volume we have only gas. The chemical potentail determines all ofthese transitions. In liquid stage we find
µl < µg
in the liquid and gas stage we findµl = µg
and finally in the gas stageµg < µl
9.1.2 Coexistence Curvep vsT
If we look at a plot of prsseure versus temperature we find thatat low temperatures, we have a line that isvery nearly to verticle hat drops down to 0 degrees. Anythingon the left we have ice, so that the chemicalpotential of the ice is less then the liquid
µice < µliq
if we now go up from 0 degrees we see that the pressure is exponential in temperature. Everything on theleft of this curve is liquid
µliq < µice µliq < µgas
and to the right of te curve we find gas, also
µgas< µliq
we also have what is known as the critical pressure and critical temperature,Pc andTc ≈ 374 degreescelsius. We also have no distinction between the liquid and the gas. To recover the previous plot wesimply need to just take a slice in our current plot to get whatare known ascoexistence curves. Theparticularly interesting things happen at the origin wherewe have have all three phases of water occuringat the same point. This is known as thetriple point of waterPt andTt . The triple point is actually used todefine the temperature scale. For water
Tt = 273.16 K = 0.010C
andPt = 4.58 mm Hg
this is used in thermometry. The line between solid and the liquid is called thefusion curve, we also havethesublimation curve, that is the line along which gas can be turned directly to a solid, and finally wehave thevaporization curve, that is the line which liquid is transformed into a gas. The fusion curve iscalled the ice line, the sublimation curve is called the frost line and the vaporization curve is caled thesteam line. These are some basic features of phase transformations.
9.1.3 Clausius-Clapeyron Equation
Lets consider two neighboring points on our coexistence curve, the first one hasPo andT0 and the secondhasP0 +dP andT0 +dT. We can say the following that the chemical potential of these two points mustbe the same. At the lower point
µg(P0,T0) = µl (P0,T0)
the higher point we find
160
µg(P0+dP,τ0+dτ) = µl (P0+dP,τ0+dτ)
we will now do a first order Taylor expansion to get
µg(P0,τ0)+
(
∂µg
∂P
)
τdP+
(
∂µg
∂τ
)
Pdτ = µl (P0,τ0)+
(
∂µl
∂P
)
τdP+
(
∂µl
∂τ
)
Pdτ
we can see that the first terms go away, we can now collect termsto get
dPdτ
=
(
∂µl∂τ
)
P−(
∂µg
∂τ
)
P(
∂µg
∂P
)
τ−(
∂µl∂P
)
τ
we can use our knowledge of the Gibbs free enery to solve this,where it was defined as
G(N,P,τ) = Nµ(P,τ)
and alsodG= µdN−σdτ+VdP
as before
V =
(
∂G∂P
)
τ,Nσ = −
(
∂G∂τ
)
N,P
and we can see that
V = N
(
∂µ∂P
)
τ,Nσ = −N
(
∂µ∂τ
)
N,P
we can now define the volume per molecule and entropy per molecule
v =VN
=
(
∂µ∂P
)
τ,N
this is the volume per molecule, and similarly
s=σN
= −(
∂µ∂τ
)
N,P
which is the entropy per molecule. We can now substitute these results into our initial result. We find
dPdτ
=sg−sl
vg−vl
now we have the pressure and temperature in a simple form. Thenumerator
sg−sl
corresponds to the increase in the entropy when one moleculechanges from a liquid to a gas. And corre-spondingly
vg−vl
161
corresponds to the increase in the volume when one molecule changes from a liquid to a gas. We can nowlook at thelatent heat of vaporization L, this is the heat required to transfomr one molecule from theliquid to the gas phase. We can relate this toτdσ
L = τ(sg−sl) = τ∆s
ordQ= τdσ
we thus find∆v = vg−vl
anddPdτ
=L
τ∆vthis is known as theClausius-Clapeyron Equation. This confirms the results from the Coexistencecurves. We can also write this in various forms that are approximations.
9.1.3.1 Approximate results
We must assume thatvg ≫ vl , an example, at 1 atmvg ≈ 1000vl . Therefore we can write
∆v = vg =vg
Ng=
τP
this assumes the ideal gas law. HencedPdτ
=L
τ∆v=
LPτ2
lets also assume that the latent heat is independent of the temperature over the range we are interested in.We can write the equation as follows
dPP
= LZ
dττ2
if we take water for example, at 0 degrees celsius, the latentheat is 2520 J/g and at 100 degrees celsius thelatent heat is 2260 J/g. We can see that this is approximately12 percent change. We now see that
lnP = −Lτ
+ lnP0
and finallyP = P0e−L/τ
this confirms the exponential behaviour of our coexistence curve. It is quite common to see this expressionwritten as
L0 = NAL (NA is Avegadros number)
gives usP = P0e−L0/RT
we can easily plot this function. The above expression is notcompletely correct but should be expressedas
lnP− lnP0 = −Lτ
+Lτ0
thus
P = P0e−L(
1τ− 1
τ0
)
162
9.1.4 Examples
9.1.4.1 CalculatedT/dP for water
If we assume that the pressure is 1 atm and the temperature is 377 K
P = 1 atm
T = 377 K
L = 2260 J/g
thusdPdT
=L0PRT2 L0 = 2260 J/g×18= 4.07×104 J/mole
alsoR= 8.31(J/K mole)thus we can see that
dTdP
=RT2
L0P= 28.4 K/atoms
9.1.4.2 Heat of vaporization for ice
We are given the following
P(−2 celsius) = 3.88 mm Hg
P(0 celsius) = 4.58 mm Hg
from this, findL of ice at -1 degrees centigrade. We will use our result
dPdT
=L0PRT2
thus
L0 =RT2
PdPdT
= −Rd lnP
d(1/T)
lets considerd lnP
d(1/T)=
d lnPdT
dTd(1/T)
we can write the followingd(1/T)
dT= − 1
T2
thusd lnP
d(1/T)= −T2d lnP
dT=
−T2
PdPdT
we just need to plug in numbers into
L0 = −Rd lnP
d(1/T)= −R
0.7−2.7×10−5 = 5.1×104 J/mole
163
9.1.5 Latent Heat and Enthalpy
If we know the specific heat of something at constant pressurewe can find the latent heat of the material.We will start with Gibbs free energy
G(N,P,τ) = Nµ(P,τ)but sometimes we have multiple phases. Each phase will have its own chemical potential. More generallywe can write the Gibbs free energy as
G = ∑j
Njµj
we can then write dow the thermodynamic identity
du= τdσ−PdV+∑j
Njdµj
now we want to talk about the transformation from a liquid to agas. Some number of molecules willchange from being a liquid to a gas
dN = dNg = −dNl
this is all valid in the coexistence curve. Therefore
du= τdσ−PdV+(µgdNg+µl dNl)
we can see that the last term vanishes in the coexistance state, thus
du= τdσ−PdV
thereforeL = τdσ = du+PdV
but we know that the enthalpy is defined as
dH = d(u+PV) = du+PdV+VdP
but we know the pressure is constant, thus
L = dH = Hg−Hl
and also
Cp = τ(
∂σ∂τ
)
P
9.2 Van der Waals Equation of State
The basic idea of this equation is to correct for both the volume and the pressure. We will look at thevolume correction first
9.2.1 Volume correction
We can reduce the effective volume to a value
V ′ = V −Nb
whereV is the volume of the box andb is some other volume. The concentration is changed by
n′ =NV ′ =
NV −Nb
in practiceb is always measured empericaly.
164
9.2.2 Pressure correction
This arises from the fact that any atom or molecule has a weak attraction to any other atom or molecule,i.e inert gas solid. This phenomenon is know as the van der Waals force. This is what is calledthe meanfield value.This refers to the idea that we neglect fluctuations. Lets suppose that we have a quantityφ(r)which is the potential energy of the interaction between thetwo atoms seperated by a distancer. We havea concentrationn, the number of atoms per unti volume. The average potential energy of all other atomsrelative to one particular atom is written as
average value of all other atoms on one atom at positionr = 0 =Z ∞
bφ(r)ndV
we have assume thatn is a constant value. The mean value “field” implies that all the atoms or moleculesare always at their “mean” position, that is to say we neglectfluctuations (usingn′ will give only a secondorder correction). We can write the integral as
−2na=Z ∞
bφ(r)ndV
thusZ ∞
bφ(r)ndV = −2a
it is minus because it is an attractive intercation, attractive force. The factor of 2 is purely for conventionand convenience. We will now use the Helmholtz free energy, thus we consider
dF = du−σdτ− τdσ
where the second and third term are zero (constant temperature and entropy). Thus
dF = du
we can therefore say
dF = du=N2
(−2na)
where the 1/2 comes from double counting. Thus
dF = −N2
Va
a is a second parameter that we have to figure out. The free energy of an ideal gas is
F = −Nτ[ln(nQ/n)+1]
we must correct this for the van der Waals gas
FdW = −Nτ[ln(nQ/n′)+1] = −Nτ[
ln
(
nQ(V −Nb)N
)
+1
]
− N2
Va
and the pressure is
p = −(
∂FdW
∂V
)
τ,N= −
(
−Nτ[
NnQ(V −Nb)
nQ
N
]
+N2
V2a
)
=Nτ
V −Nb− N2
V2a
165
we finally get the following(
p+N2
V2a
)
(V −Nb) = Nτ
this is the van der Waals equation for a gas. If you think of theinteractions of atoms or molecules you findthat at very low seperations there is a repulsive force whichis usually modeled as≈ 1/r12. On the otherhand if you start pulling the hydrogen molecule apart there will be an attraction given by 1/r6, which isthe van der Waals term. Where the 1/r6 can be looked as the 1/V2 term arising from the van der Waalsforce. We can now use this equation to predict critical behavior.
9.2.3 Finding critical point
If we look at thep versusV plot we can see that we have three critical points for phase transitions. Thereare a series of isotherms that can also be seen in the plot. Aobe some critical point we get what is calleda fluid and this happens at the critical isotherm which occursat τC. These isotherms occur on the vaporstaturation curve. The thing that will enable us to find the critical point is that
(
∂p∂V
)
τC
= 0
(
∂2p∂V2
)
τC
= 0
knowing that
p =Nτ
V −Nb− N2
V2a
thus(
∂p∂V
)
= − Nτ(V −Nb)2 +2
N2
V3a = 0 ⇒ 2NaV3 =
τ(V −Nb)2
and(
∂2p∂V2
)
= − 2Nτ(V −Nb)3 +6
N2
V4a = 0 ⇒ 3NaV4 =
τ(V −Nb)3
If we devide the equations we find2VC
3= VC−Nb
thus we have just found the critical volume. We get the following result
VC = 3Nb (9.1)
this is by far the most accurate way to get these parameters (a andb). If we take the previous expressionand insert into the first equation we find
τC =2Na
27N3b3(2Nb)2 =827
ab
(9.2)
we also want to findpC from the van der Waals equation
pC =4Na27b
12Nb
− N2a9N2b2 =
427
ab2 −
327
ab2 =
a27b2 (9.3)
henceVC andτC definea andb. We want to eliminate the three parametersa, b andN from the van derWaals equations. If we start with
(
p+N2
V2a
)
(V −Nb) = Nτ
166
we can write(
p+
(
VC
3b
)2 27b2pC
V2
)
(
V −VC
3
)
=VC
3bτ
we still need to get rid ofb, we can devide 9.3 with 9.2 we get
pC
τC=
a27b2
27b8a
=18b
thus we find(
p+
(
VC
3b
)2 27b2pC
V2
)
(
V − VC
3
)
=8VC
3τpC
τC
this is an equation that involvesp,V andτ andpC,VC andτC. We can divide through by the productpCVC
thus(
ppC
+3
(
VC
V
)2)
(
VVC
− 13
)
=83
ττC
where we are only left with the ratios of all these quantities. We can now introduce
p =ppC
V =VVC
τ =τ
τC
which are rthe reduced values, thus we can write
(
p+3
V
)(
V − 13
)
=83
τ
two gases with the same values of ˆp V andτ are said to be in corresponding states, i.e all gases look thesame in terms of these three variables. Lets go back to the previous expression and write it as
(pV +3)
(
V − 13
)
=83
τV2
this is a general cubic expression inV. This expression can be solved numerically, but in general acubicfunction gives an S shaped curve with two turning points. If you decrease the pressure and increase thevolume we come to a metastable state, and likewise if you decrease the volume and increase the pressurealso allow you to reach a metastable state. The turning pointfor the first case is called superheating andthe second case is supercooling. This formalism breaks downwhenV < 1/3. If we consider the line forpA (lower left minimum) we can write the Gibbs freen energy as
dG= −σdτ+Vdp+µdN
but at constantτ andN we finddG= Vdp
so as we go from the left minimum to the right maximum then the total change from this is given as
∆G =Z B
AdG= 0
167
we can write down the integral in the following way
Z B
AVdp =
[
Z D
A+
Z C
D+
Z E
C+
Z B
E
]
Vdp
=
[(
Z C
D−
Z A
E
)
−(
Z E
B−
Z E
C
)]
Vdp
where the first two integrals is the area of ADC and the second set of integrals is the area of CBE and thisis all equal to 0. We define ˆp so that the areas are equal.
9.3 Nucleation of a Liquid Drop
This process is concerned with how rain is created, i.e how doyou build liquid drops from gases. We aretalking aboutmetastability. Lets us consider
∆µ= µg−µl
where these refer to the “bulk” properties. If∆µ> 0 then gas condenses to liquids. For a drop, the surfacetension (energy) increases the chemical potential for the liquid and may prevent condensation and thisleads to the concept of thecritical radius of the drop. If the drop is above this radius it can grow but if itis below this area than it shrinks
R > Rc grows
R < Rc shrinks
recall that the Gibbs free energy is given asG = Nµ
and the change in the Gibbs free energy is given as
∆G = Gl −Gg =43
πR3nl (µl −µg)+4πR2γ
whereγ = G/A is the Gibbs free energy per unit area. The first term dominates for large radius and thesecond term dominates for small radius. To find the critical radius we take the derivative
d∆GdR
= −4πR2nl∆µ+8πRγ = 0
thus the critical radius is given as
Rc =2γ
nl ∆µ
if we take the second derivative
d2∆GdR2 = −8πRnl∆µ−8πγ = −16πγ+8πγ = −8πγ < 1
thus this is a maximum. What is the height of the barrier?
∆Gb = −43
πR3cnl ∆µ+4πR2
cγ
168
substituting our expression for the critical radius we find
∆G = −43
π8γ3
n3l ∆µ3
nl ∆µ+4π4γ2
n2l ∆µ2
γ
=
(
16− 323
)
πγ3
n2l ∆µ2
∆G =16πγ3
3n2l ∆µ2
Lets look at what the critical radius for a water droplet is. If we have a water droplet which is sorroundedby vapor which applies a pressurep> pbulk andp(τ) exceedspbulk(τ), where the bulk caseµl = µg. Recallthat
µ= τ ln
(
nnQ
)
= τ ln(Ap)
since we know thatp ∝ n at fixedτ. Thus
∆µ= µp−µp(bulk) = τ ln(Ap)− τ ln(Apbulk) = τ ln
(
ppbulk
)
we have related the change in chemical potential to the ratioof these two pressures. Putting some numbersfor water
τ = 300 Kp
pbulk= 1.1 nl =
6×1023
18=
1023
3cm3 =
1029
3m3 γ = 7.2×10−2J m−2
thus
Rc =2γ
nl ∆µ=
14.4×10−2(
1029
3
)
(1.4×10−23×300ln(1.1))m = 10−8m
9.4 Summary
The Clausius-Claperyin equationdpdτ
=L
τ∆Vwhere
L =Lt ·ht
moleculeand the pressure is given as
P = P0e− L0
R
(
1T − 1
T0
)
and
L0 =Lt ·htmole
and the van der Waals equation is given as
(
p+N2
V2a
)
(V −Nb) = Nτ
169
the critical point is given as
Vc = 3Nb pc =a
27b2 τc =8a27b
and corresponding states(
p+3
V
)(
V − 13
)
=83
τ
9.5 Problems
Problem # 1 Entropy, energy, and enthalpy of van der Waals gas
a) Show that the entropy of the van der Waals gas is
σ = N
[
ln
[
nQ(V −Nb)N
]
+52
]
We know that the free energy of the gas, with the van der Waals correction is given as
FdW = −Nτ[
ln
(
nQ(V −Nb)N
)
+1
]
− N2
Va
and we know that the entropy is defined as
σ = −(
∂F∂τ
)
V= − ∂
∂τ
[
−Nτ
[
ln
(
kτ3/2(V −Nb)N
)
+1
]
− N2
Va
]
wherek contains all of the constants that come with the quantum concentration. We know that
∂∂τ
(AB) = B∂A∂τ
+A∂B∂τ
if we let
A = −Nτ B =32
ln(kτ)+ ln(V −Nb)− lnN+1
thus we find
∂∂τ
(AB) = −N
[
32
ln(kτ)+ ln(V −Nb)− lnN+1
]
−Nτ∂∂τ
(
32
ln(kτ))
= −N
[
32
ln(kτ)+ ln(V −Nb)− lnN+1+32
]
= −N
[
ln
(
nQ(V −Nb)N
)
+52
]
thus the entropy is given as
σ = N
[
ln
[
nQ(V −Nb)N
]
+52
]
b) Show that the energy is
U =32
Nτ−N2 aV
170
We know that the total energy is given asU = F + τσ
thus
U = −Nτ[
ln
(
nQ(V −Nb)N
)
+1
]
− N2
Va+Nτ
[
ln
[
nQ(V −Nb)N
]
+52
]
=32
Nτ−N2 aV
thus the total energy is given as
U =32
Nτ− N2
Va
c) Show that the enthalpyH = U + pV is
H(τ,V) =52
Nτ+N2bτV
−2N2 aV
H(τ, p) =52
Nτ+Nbp−2Napτ
All results are given to first order in the van der Waals correctio termsa,b.
If we start with(
p+N2
V2a
)
(V −Nb) = Nτ
expanding this equation gives
pV− pNb+N2
Va− N3
V2ab= Nτ
thus
pV = pNb− N2
Va+
N3
V2ab+Nτ
since all the results are given to first order allows us to simplify
pV = pNb− N2
Va+Nτ
thus the enthalpy is terms of the volume gives
H(τ,V) = U + pV =52
Nτ+N2bτV
−2N2 aV
and since we know thatp = Nτ/V ⇒V = Nτ/p and alsoN/V = p/τ, thus the enthalpy in terms of thepressure give
H(τ, p) =52
Nτ+Nbp−2Napτ
Problem # 2 Latent heat of magnesium
171
In the temperature range 700 to 730 K, the vapor pressure of magnesium can be approximately representedby
log10 p = −7,500T
+8.6
(p in mmHg, T in K). What is the molar latent heat of sublimation of magnesium in this temperaturerange?
We know that the latent heat of is given as
L = τ(s1−s2)
wheres1 ands2 are the entropy for the specific phase of the subtance, thus the latent heat of fusion (solidto liquid) is
L f = τ(sl −ss)
the latent heat of vaporization is (liquid to gas)
Lv = τ(sg−sl)
and the latent heat of sublimation (solid to gas)
Ls = τ(sg−ss)
we also know that
ln p = −L0
τ+constant
but we are given
log10 p = −7,500T
+8.6
so it is just a matter of comparing these two equation and doing a change of base. We also know that
ln p = loge p =log10 ploge
thuslog10 ploge
= ln p
and we find
log10 p = − 7,500T loge
+8.6
loge
and we can see that the latent heat is given by
L0 =7,500kB
loge= 2.38×10−19 J
K= 1.43×105 J
mole
and so
L0 = 1.43×105 Jmole
Problem # 3 Latent heat of neon
172
The latent heat of fusion and the latent heat of vaporizationof neon at the triple point have values335J/mole and 1810 J/mole respectively. The temperature ofthe triple point is 24.57 K. What is the latentheat of sublimation of neon at the triple point? What is the change in entropy when one mole of liquidneon at the triple point is vaporized?
Since we know that the latent heat of fusion (solid to liquid)is
L f = τ(sl −ss)
the latent heat of vaporization is (liquid to gas)
Lv = τ(sg−sl)
and the latent heat of sublimation (solid to gas)
Ls = τ(sg−ss)
we can see that
ss = sl −L f
τsg =
Lv
τ+sl
thus we can see that the latent heat of sublimation is
Ls = Lv +L f
thus we find that the laten heat of sublimation is
Ls = 2145J
mole
and the change in entropy for vaporization is given by
sg−sl =Lv
τ=
Lv
kBT×mole
which is given by
∆σ = 5.33×1024
173
Chapter 10
Kinetic Theory
1. Maxwell distribution of speeds
2. Mean free path
3. Transport: Diffusion, thermal conductivity, self-diffusion
4. “Ballistic regime”- Knudsen gas
10.1 Maxwell distribution
We are only consrened with the magnitude of teh velocities and not direction. We must consider thenotion of quantized energy in a box. We can combine this with the probability that any one of these statesis occupied, i.e combine quantum energy levels withf (εl), recall
f (εl) =n
nQe−εl/τ where εl =
~2
2m
(
πlL
)2
we will confine our system to have only spin 0 atoms. We want to express the number of atoms withquantum numbers betweenl andl +dl this goes as
N(εl) = 1× 18×4πl2×dl× f (εl ) =
π2
l2 nnQ
e−εl/τdl
where 1 is for the spin 0 and184π2dl number of states in shell of octant we want to convert from quan-tum numbers(l ) to speed(v), thus
εl =12
mv2 =~
2
2m
(
πlL
)2
this is the important step that takes you from quantum numbers into speed
v2 =~
2
m2
π2l2
L2 l =MLπ~
v
if we say that there areN molecules in a volumeV and lets introduce a probability that a particle has avelocity in a particular rangep(v) is the probability that molecules have speeds betweenv andv+dv, thus
Np(v)dv=12
πl2 nnQ
e−εl/τ dldv
dv
174
wheredldv
=MLπ~
putting this together we have
Np(v)dv=12
πn
nQ
(
MLπ~
)2
v2e−Mv2/2τ(
MLπ~
)
dv
lets recalln
nQ=
NL3
(
2π~2
Mτ
)3/2
and putting everythign together we find
Np(v)dv=π2
NL3
(
2π~2
Mτ
)3/2(MLπ~
)3
v2e−Mv2/2τdv
we can see that a lot of terms cancel
p(v) = 4π(
M2πτ
)3/2
v2e−Mv2/2τ
this is the famous Maxwell distribution for speeds of a classical ideal gas. We can now derive the RMSspeed for the distribution.
Three speed
The RMS speed is defined as follows
〈v2〉 =
Z ∞
0v2p(v)dv= 4π
(
M2πτ
)3/2 Z ∞
0e−Mv2/2τv4dv
where we can write
4π(
M2πτ
)3/2
=4π2
(
M2τ
)3/2
thus
〈v2〉 =4π2
(
M2τ
)3/2 Z ∞
0e−Mv2/2τv4dv
we can sety = Mv2/2τ thus
v =
(
2τM
)1/2
y1/2 dv=
(
2τM
)1/2 12
y−1/2
and
v4dv=12
(
2τM
)5/2
y3/2dy
175
and finally we find
〈v2〉 =4
π1/2
(
M2τ
)3/2(2τM
)5/2 Z ∞
0e−yy3/2dy
=32
2τM
=3τM
therefore√
〈v2〉 =
√
3τM
this is the same result we derived in the equipartition theorem, i.e
12
mv2 =32
τ
Lets now calculatethe mean speedwhich is not zero, but we know that the maen velocity is zero. We canwrite
〈v〉 =Z ∞
0vp(v)dv=
4π2
(
M2τ
)3/2 Z ∞
0e−Mv2/2τv3dv
making the same substitution as before gives
v3dv=12
(
2τM
)2
ydy
plugging this in gives
〈v〉 =4
π1/2
(
M2τ
)3/2(2τM
)2 Z ∞
0e−yydy
which yields
〈v〉 =
(
8τπM
)1/2
thus this is still of the formτ/M, thus the maen speed is slightly different than the RMS speed. We cannow derive what themost probable speed. We can see that at low speeds, the distribution is quadratic, butwith increasing speed the exponential term will dominate. The maximum will give us the most probablespeed. The most probable speed can be derived by differentiating the previous expression
p(v) = Av2e−Mv2/2τ
wheredp(v)
dv= A
(
2v− Mv3
τ
)
e−Mv/2τ
thus
v2MP =
2τM
and finally
vMP =
√
2τM
this are the three speeds that gases can have. We can normalize to findv
vMP=
v√
2τ/M
depending exactly on what the problem is, you will have to utilize one of these expression.
176
10.2 Mean Free Path
The mean free path is the average distance traveled by an atombetween collisions. It is almost alwaysgiven asl . We can replace the fact that atoms of like kind have a diameter d which are all moving around indifferent directions. We replace the previous picture by assuming that one atom has a diameter 2d (superatom) and all other atoms are points. We can consider our super atom will move a distancel in a cylinderwith length l . On the average there is one atom in the cylinder of volumeV = lπd2, that is to say sincethere is one atom in the cylinder
lπd2 = n = 1
wheren is the number of atoms per unit volume, thus
l =1
πnd2
how big is this number?
10.2.1 Estimate ofl
Lets first assume that we are in STP conditions. We also have the Loschmidt number
n =Avegadros number
molar volume at 0 celsius and atm. pressure=
6.02×1023
22.4×10−3m3 = 2.69×1025m−3
we can now make an estimate of the mean free path
l =1
π×2.69×1025m−3×2×10−10m2 ≈ 300 nm
how often does this collision occure? Themean free timeis found by
t =lv
=3×10−7 m
103 m/s≈ 0.3 ns
lets take another limit, for example an ultra high vacuum system
p∼ 10−13 atm
thusl ∼ 3×10−7 m×1013 ∼ 3000 km
the mean free path becomes really large under low pressures.This forces very different behaviors forgases under different conditions. The mean free time in thissystem is
t ∼ 0.3 ns×1013∼ 1 hr
when the MFP> dimensions of the box is called the “ballistic” regime or “Knudsen” regime.
10.3 Transport
We will now talk about how atoms move through a medium this is know astransport . We are assumingthat we have particle diffusion when there is a gradient of concentration. This is in analogy to Ohm’s law.
177
10.3.1 Particle diffusion in a concentration gradient
Lets consider a resevoir with two cavities, a left and right cavity. Lets assume that the left cavity has ahigher concentration than the right, thus there is a steady flow of particles from the left to the right cavity.We want to figure out the rate of motion of these particles depend on the speed and the mean free path.There is a net flow from higher concetration to the lower concentrationµl > µr . The flux of particles isdefined byFick’s law, which says
Jzn = −D
dndz
whereD is the diffusive constant (diffusivity), the minus sign comes from the fact that the gradient ispositive to the left but particles flow to the right. The flow isfrom highn to low n. We have definedlz asthe mean free path along thez axis andvz is thez component of the average velocity. At the planez wehave the flux in the +z direction is given by
12
n(z− lz)vz
and the flux in the -z component is12
n(z+ lz)vz
and the net flux is the difference between these two components
Jzn =
12
[n(z− lz)−n(z+ lz)] vz
we can write the term on the left, with a Taylor expansion
n(z+ lz) ≈ n(z)+ lzdndz
and likewise for the other term, thus
Jzn = −dn
dzvzlz
we have not taken into account for the fact that not all these atoms travel in thez direction, thus we musttake an angular average to account for this. Thus the projection on thezaxis is given by
vz = vcosθ
andlz = l cosθ
wherelz is the projection of the MFP on thez axis. Recall that the solid angle is defined as
dΩ = 2πsinθdθ
we want
〈vzlz〉 =vl2π
R π/20 cos2θsinθdθ
2πR π/2
0 sinθdθwhere the denominator is for normalization. Thus we can see
〈vzlz〉 =vl
R π/20 cos2θsinθdθR π/2
0 sinθdθ= vl
[
−13 cos3θ−cosθ
]π/2
0
=vl3
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thus the diffusion constant is
D =vl3
how big is this constant? Lets take a gas at STP. Recall that ¯v≈ 105 cm/s andl ∼ 3×10−5 cm and thusD∼ 1 cm2/s. This is the basic problem of atoms diffusing in a gradient and thus this increases the entropy.
10.4 Thermal Conductivity
We want to replace the scenerio where we have a concentrationgradient with one that has a temperaturegradient. Imagine that we have two reseviors with temperature τ1 andτ2 whereτ1 > τ2. Thus we nowhaveFourier’s law
Jzµ = −K
∂τ∂z
whereK is known as the thermal conductivity. We want to write this asfollows
Jzu = −D
∂ρu
∂z
whereρu is the energy density, thus this is an exact analogy from whatwe wrote before, we can write
Jzu = −D
∂ρu
∂τ∂τ∂z
where we can see that∂ρu
∂τ= CV
thus
Jzu = −DCV
∂τ∂z
and hence
K = DCV =vl3
CV
this result shows us that the thermal conductivity is independent of the pressure of the gas,CV ∝ n andl ∝ 1/n and henceK is independent ofn and hence the pressure. This is provided that the mean free pathis still much less then the dimensions of the box. How big is it? At STP, the heat capacity is
CV =32
kBn∼ 32×3×1019cm−3×1.4×10−23 J/K∼ 10−3 J/K cm−3
and thusK = CVD ∼ 10−3 J/Kcm3×1 cm2/s∼ 1 mW/cmK
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