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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 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 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
∂σ
=
τ
∂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
∂σ
− =
τ
∂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
6. Ideal classical gas
7. Quantum Gases: They come in two flavors, Fermi gas and Bose gas. The Fermi gas is for fermions
and 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 this g of a
system with given energy which is equal to the number of quantum states with that energy ε. e.g Particle
in 3D box
nx ny nz
1 1 1
1 1 2
1 2 1
2 1 1
the first energy state is non degenerate g = 1 the second energy state is three-fold degenerate g = 3.
Quantum states of a one-particle are called “orbitals”. For Non interacting Spins
N= ↑
↑ ↑ ↓
↑ ↑
↓ ↑
the total number of arrangements is given by 2N , the total magnetic moment is given by M = Nm for all
spins that are ↑ and M = −Nm for all spins that are ↓. How many ways can we get M = (N − 2)m (flip
over any one spin). We will call g the number of ways in which we can get a particular value of a
magnetic moment. Lets define the following
N
N
+ s ↑ n↓ =
−s ↓
n↑ =
2
2
where N is even and M = (n↑ − n↓ )m = 2sm where the 2s = spin excess = n↑ − n↓ . If we assume each
spin has a probability P↑ of being ↑ and a probability of P↓ of being ↓ . we also know that P↑ + P↓ = 1. In
the presence of a magnetic field ↑, P↑ > P↓ . The probability of a configuartion with spin excess 2s is
N/2−s
N/2+s
. The probability of finding a magnetic moment M = (2s)m is given by the following
and P↓
P↑
N/2+s N/2−s
P↓
P(N, s) = g(N, s)P↑
we need to find what the multiplicity g(N, s) is. In how many ways can we distribute n↑ = N/2 + s spins
over N places. we can assume that they are distinguishable spins. So we can assume that the first spin can
be on any one of the N sides.
3
• 1st spin can be in any one of the N 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 about
the 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 the n↑ magnets
• 2nd of spin up could have been any one of the n↑ − 1 magnets
• nth of spin up could have been any one of the 1 magnets
Thus we have overcounted by n↑ (n↑ − 1)(n↑ − 2)... − 1 = n↑ !. Thus
g(N, s) =
N!
(N − n↑ )!n↑ !
This is known as the multiplicity function; it is the number of states having the same value s. If we
replace
N
n↑ = + s
2
yields
N!
N
g(N, s) = N
2 −s ! 2 +s !
This is the number of ways of achieving a spin excess of 2s. Hence
N!
N/2+s N/2−s
N
P↑
P(N, s) = N
P↓
−
s
!
+
s
!
2
2
For the moment lets assume that we dont apply any field B = 0then P↑ = P↓ = 12 , thus we find
P(N, s) =
If we have the following spins
N
2
N!
1
N
N
−s ! 2 +s ! 2
N spins ↑ ↑
↓ ↑
↑ ↓
the probability is given as
P(N, s) =
g(N, s)
2s = spin excess
2N
where
g(N, s) =
N
2
N!
N
−s ! 2 +s !
4
1.2 Gaussian Distribution
we first need to assume
N ≫ 1, N even
|s| ≪ N P↑ = P↓ =
1
2
we will use ln as the natural logrythim, thus
N
N
ln g(N, s) = ln N! − ln
+ s ! − ln
−s !
2
2
working on this expression yields
N
N
N
N
+ s ! = 1, 2, 3...
+ 1 , ...
+s
2
2
2
2
N
N
N
!
+ 1 , ...
+s
=
2
2
2
s
N
N
N
ln
! + ∑ ln
+ s ! = ln
+k
2
2
2
k=1
and now we need to do it for the other one
1, 2, 3, .. N2 − s N2 − s + 1 , ... N2
N
−s
=
N
N
2
2 .... 2 − s + 1
N
2 !
=
N/2...... N2 − s + 1
s
N
N
N
!0 − ∑ ln
− s ! = ln
−k+1
ln
2
2
2
k=1
hence
#
"
s
1 + 2k
N
N
N
N
ln
! + ∑ ln
+ s ! + ln
− s ! = 2 ln
2
2k
2
2
2
+
1
−
k=1
N
N
and since we know that s ≪ N and also k ≪ N, we can expand the expression
s s 2k
2k 2
2k 2
2k
≈ ∑
− − +
∑ ln 1 + N − ln 1 − N + N
N N
k=1
k=1 N
1
4 s
k−
≈
∑
N k=1
2
when we add all these terms together and find that there are s terms, thus the sum is given by
2s2
1
4 s
=
k
−
∑
N k=1
2
N
putting this all together yields
2s2
N
ln g(N, s) = ln N! − 2 ln
!−
2
N
5
thus
2 #
g(N, s) N2 !
2s2
ln
=−
N!
N
"
and
g(N, s) =
N
2
and to summarize
2 /N
g(N, s) = g(N, 0)e−2s
2
N!
N e−2s /N
! 2 !
where g(N, 0) =
N
2
we also know that this function falls of as 1/e when
2s2
= 1, s =
N
1/2
N
2
s
=
N
1
2N
N!
N
! 2 !
1/2
thus, if N is very large, the peeking is exceedingly sharp. For large N, thermodynamic quantities (heat
capacity,paramagnetic susceptibility are very well defined. To take an example, lets assume that
N = 1020 (1 cm3 air)
s
≈ 10−11
N
How to evaluate N!? There is a useful approximation called Sterling’s approximation, it is defined as
for large N N! ≈ (2πN)1/2 N N e−N
we need to work with this until we make it look like the multiplicity g(N, s)
1
ln N! = N ln N − N + ln(2πN)
2
note that the error is less the 1% for N = 10 and it gets much better for higher N. Lets use this to evaluate
g(N, 0)
#
"
1
N!
N N N 1
ln g(N, 0) = ln N N = N ln N − N + ln(2πN) − 2
ln − + ln(πN)
2
2 2 2 2
2 ! 2 !
let us do
1
1
2
1
2
ln g(N, 0) = N ln 2 + ln(2πN) − ln(πN) = N ln 2 + ln
2
2
2
πN
or more compactly
"
ln g(N, 0) = ln 2N
2
πN
1/2 #
and finally
g(N, 0) = 2
N
6
2
πN
1/2
1.2.1 Notes on The Gaussian
2 /2σ2
g(x, σ) = Ae−(x−x̄)
where
x̄ =
=
6=
σ =
mean value
0 for P↑ = P↓
0 for P↑ 6= P↓
standard deviation
where σ is used for error analysis, this is a method of expressing the confidence 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 as x → ∞
• Events are statistically independent
1.4 Mean Values
Suppose that function f (s) has a probability of P(s) of occuring. Then the mean value is
h f i = f¯ = ∑ f (s)P(s) where
s
∑ p(s) = 1
s
e.g, the toss of a die
6
n̄ =
1
1
∑ np(n) = (1 + 2 + 3 + 4 + 5 + 6) × 6 = 3 2
n=1
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
g(N, s)
h f i = f¯ = ∑ f (s) N
2
s
7
if we let
f (s) = s2
then
hs2 i =
6 hsi2
but
2
hs i =
=
if we let
2s2
x =
N
2
and so we find
2
hs i =
now lets calculate the spin excess
2
πN
∑s 2
2 N
s
2
πN
2
πN
1/2 Z
1/2
∞
−∞
2 /N
e−2s
2 /N
s2 e−2s
1
2N
ds
1/2
1/2
N
N
x ds =
dx
s=
2
2
1/2
N
2
1/2 Z ∞
N
2
N
x2 e−x dx =
2
4
−∞
h(2s)2i = N mean square
√
h(2s)2i1/2 =
N root mean square
to find the fractional rms spin excess
1
h(2s)2i
=√
N
N
1.5 Summary
1. Multiplicity for N spins 1/2 with spin excess 2s is given by
N!
N
g(N, s) = N
+
s
!
−
s
!
2
2
but for N ≫ 1, and s ≪ N we can write
2 /N
g(N, .s) = g(N, 0)e−2s
where
1/2
N!
2
N
g(N, 0) = N N = 2
πN
2 ! 2 !
and using Sterlings approximation
1
ln N! = N ln N − N + ln(2πN)
2
2. the Mean value
h f i = ∑ f (s)P(s), where
s
and the spin system
1
h(2s)2i
=√
N
N
8
∑ P(s) = 1
s
1.6 Problems and Solutions
Problem # 1
A penny is tossed 400 times. Use the Gaussian distribution to find 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) =
2
πN
1/2
2 /N
2N e−2s
And the probability is given as
P(N, s) =
g(N, s)
2N
and since we know that
N = 400 s = x − x̄ = 215 − 200 = 15
we find that the probability of getting 215 heads is given as
P(400, 15) =
2
π(400)
1/2
9
2 /400
e−2(15)
≈ 1.3%
Problem # 2
The 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 during
the peak use hour of the day, each campus phone is used to make a single two-minute call at a random
time. Find the minimum number of telephone lines required so that at most only 1% of callers fail to have
immediate access to a line. (Hint: approximate the distribution by a Gaussian distribution, and use tables
or your computer.).
Method 1
Since we know that in any 2 given minutes, the average number of people using a telephone line is
given by x̄ = hxi = N p = 2000
30 . We find that this distribution can be modeled as a Gaussian distribution of
the form
The distribution plotted above is given by
p(x) = √
where
1
2 /2σ2
2πσ2
e−(x−x̄)
σ2 = N p(1 − p)
where p is the probability that a phone line will be used which is
p=
1
30
N = 2000
so all we have to do is integrate this function to some value of x that will yield 99%, thus
p(x < xl ) = √
1
2πσ2
10
Z x
0
2 /2σ2
e−(x−x̄)
dx
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% of the people are not able to make a
call.
N = 86 lines
Method 2
If we say that n↑ are the number of phone lines being used
N
+s
2
n↑ =
N
−s
2
n↓ =
and if we let
N
+s = N
2
and we also know that the probability is given as
n↑ =
s=
N
2
N/2+s N/2−s
P↓
P(N, s) = g(N, s)P↑
where the probability of the number of lines being used is given by
P↑ =
2 min 2000 lines 200
=
60 min N lines
3N
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) = 2
but since we know that s =
N
2
N
2
πN
1/2
2 /N
e−2s
we get
g(N, N/2) = 2
N
2
πN
1/2
200
3N
N/2+s e−N/2
thus we find that the probability is given by
P(N, s) = 2
given that s =
N
2
N
2
πN
1/2
−2s2 /N
e
3N − 200
3N
this simplifies into
P(N, N/2) = 2
N
2
πN
1/2
11
−N/2
e
200
3N
N
N/2−s
(1.1)
and since we know that the probability needs to be 0.01% we can just write this as
2
N
2
πN
1/2
−N/2
e
200
3N
N
= 0.01
This can only be solved numerically, the numerical solution to this is
N = 83.0125 ≈ 84 lines
Thus we know that
P(83, 83/2) = 0.01
Method 3
If we model this as a binomial distribution
p(n) =
N!
pn (1 − p)N−n
(N − n)!n!
to find the minimum number of phone lines that will be needed so that fewer than 1 % of the callers fail to
get thru requires us to take a sum
N
p(n < nl ) =
N!
pnl (1 − p)N−nl < 0.01
nl +1 (N − nl )!nl !
∑
where nl is the minimum number of phone lines required to satisfy this criterion. This can easily be put
into Mathematica, but I dont have it so I will quote the solution as
nl = 86
Method 4
If we look at the following function
p(x) = √
1
2πσ2
Z x
0
2 /2σ2
e−(x−x̄)
dx
We can see that this function resembles the error function
2
erf(x) = √
π
Z x
0
2
e−u du
and the soultion for x 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 looking
for an x that will give us 99%. Thus we know that
u = erf−1 (erf(x)) = erf−1 (0.99) = 1.822
but we know that our u is simply
x − x̄
√ = 1.822
2σ
12
since we know that
σ=
thus we find
p
N p(1 − p) x̄ =
x=
√
2000
30
2σ(1.822) + x̄
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 # 3
The binomial distribution can be written in the form
N
P(N, n) =
pn (1 − p)N−n
n
N
n
=
N!
n!(N − n)!
where p is the probability that some single event occurs.
Assume that p ≪ 1, and N ≫ n. Show that P(N, n) can be written in the approximate form
λn e−λ
n!
where λ = N p. This is the well-known Poisson distribution (first derived to study the death rate from
horse kicks in the French army!). Sketch P(N, n) vs. n for λ < 1 and λ > 1.
P(N, n) =
Since we know that
P(N, n) =
where the binomial coefficient is
N
n
=
N!
(N − n)!n!
N
n
pn (1 − p)N−n
or using Newton’s generalized binomial theorem
N(N − 1)(N − 2)...(N − (n − 1))
N
=
n
n!
but since we know that N ≫ n this is simply
N
n
=
Nn
n!
thus we can write our original expression as
P(N, n) =
we can see that that
(1 − p)
N−n
(N p)n
(1 − p)N−n
n!
∞
=
∑
k=0
13
N −n
k
(−p)k
but we can see that
(N − n)(N − n − 1)(N − n − 2)...(N − n − (k − 1)).. N k
N −n
=
=
k
k!
k!
since we know that N ≫ n. Thus we find
(1 − p)N−n =
∞
(−N p)k
∑ k! = e−(N p)
k=0
thus we find
(N p)n −(N p)
e
n!
but if we let N p = λ then we recover what we were looking for
P(N, n) =
λn −λ
P(N, n) = e
n!
and the Poisson curves are given by
We can see from the plots that the Poisson distribution is well defined for large values of N, in actuallity
the Binomial distribution becomes the Poisson distribution for large value of N.
Problem # 4 The Meaning of Never
It 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 it
gives a misleading conclusion about very, very large numbers. Could all the monkeys in the world have
typed 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. We
suppose that a monkey can hit 10 typewriter keys per second. A typewriter may have 44 keys; we accept
lowercase letters in place of capital letters. Asuuming that Shalespear’s Hamlet has 105 characters, will
the monkeys hit upon Hamlet?
14
a) Show that the probability that any given sequence of 105 characters typed at random will come out in
the correct sequence (the sequence of Hamlet) is of the order
1
44
100000
= 10−164345
where we have used log10 44 = 1.64345.
Since we know that there are a total of 44 keys then we know that the probability that a key will land upon
the right place is given by
1
p(1) =
44
but since we know that there are a total of 105 characters in total for Hamlet we know that the total
probability of this happening is given by
p(N) =
1
44
N
but since we know that N = 105 this is simply
p(N) =
1
44
100000
but we know that
log10 p(N) = 100000 log10
1
44
≈ −164, 345
thus
p(N) = 10−164345
b) Show that the probability that a monkey − Hamlet will be typed in the age of the universe is approximately 10−164316 . The probability of Hamlet 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 literary
production of the monkeys
Since we know that there is
Nmonkeys = 1010
tuniverse = 1018 s
10 keys
Nkeys /s =
s monkey
thus the total number of keys that these monkeys can type in the total age of the universe is
10 keys
= 1029 keys
Nkeys = 1010 monkeys × 1018 s ×
s monkey
the probability that a monkey − Hamlet will 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 that N = 10 and that the spin excess 2s = n↑ − n↓ = 0. We also know that the probability of this
two-state system is given by
g(N, s)
P(N, s) =
2N
we also know that
N!
N
g(N, s) = N
2 +s ! 2 −s !
thus
g(10, 0) =
10!
5!5!
and the probability is given as
P(10, 0) =
10! 1
= 0.25
5!5! 210
b) You flip 1000 coins. What is the probability of obtaining exactly 500 heads and 500 tails?
We know that N = 1000 and the spin excess is 2s = 0. Since we also know that s ≪ N allows us to make
teh Sterling approximation
2
g(N, s) = g(N, 0)e−2s /N
where
g(N, 0) = 2
N
2
πN
1/2
and so the probability is given by
g(N, s)
=
P(N, s) =
2N
thus we find
P(1000, 0) =
r
r
2
πN
2
= 0.025
1000 × π
c) Why is the answer to a) more than b), yet we say the multiplicity function sharpens with increasing N.
The probability of getting exactly zero “spin excess”, i.e 50 % heads √
and 50 % tails decreases as N
increases. But the probability of being very close to s = 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 the specifications of a system. e.g lets take a cubic
box: for 1 particle the energy of the state is identified by three quantum number, nx , ny , and nz .
• 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 quantum states
• The total energy of a particle must 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 these
systems. However, in statistical mechanics we consider an ensemble average. The ensemble contains g
identical systems. One for each accesible state, where g is the multiplicity. We then average over the
ensemble 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) =
4!
N!
=
=4
n↑ !n↓ ! 3!1!
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 state k with
spin excess 2 is
2 = p(k)
where
∑ p(k) = 1
k
Lets suppose that we have a parameter x that takes the value x(k) when the system is i the state k, this
could be the magnetic suceptibility or the heat capacity. Then the expectation value of x is
hxi = ∑ p(k)x(k)
ensemble average
k
It is reasonable to suppose that the time average is the same as the ensemble average, this is known as
Ergodic 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, or
Maxwell’s equations. All we can do is to build a theory based on this postulate and then we can compare the results with the experiments. If we get predictions that are correct, than we ca say our postulate
is justified “a posteori” (after testing). Thus
p(k) =
18
1
g
and thus
hxi =
1
x(k)
g∑
k
Example
Lets suppose that we have the four magnetic spins all with different electric dipole moments. Lets let x
be the electric dipole moment d, 2d, 3d, and 4d. For the four systems we have the following net electric
dipole moments
↓ ↑
↑ ↑
↑ ↓
↑ ↑
↑ ↑
↓ ↑
↑ ↑
↑ ↓
−d + 2d + 3d + 4d
2 − 2d + 3d + 4d
d + 2d − 3d + 4d
d + 2d + 3d − 4d
So the expectation value is
hdi =
=
=
=
=
8d
6d
4d
2d
∑ x(k) 8d + 6d + 4d + 2d
=
= 5d
g
4
2.4 Two Systems in Thermal Contact: Most Probable Configuration
(MPC)
As an example, consider spin system. Lets suppose that there is a magnetic field B
N1, u1 , s1 system 1
N2 , u2 , s2 system 2
where S is the spin excess, this is a closed system with a total energy ET . When we put this two together
we 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 energy
u = −2smB = −2(s1 + s2 )mB = −2(s′1 + s′2 )mB
We also need to assume that N1 < N2 , and thus we have the following constraint
|s1 | ≤
N1
or − N1 ≤ 2w ≤ N1
2
lets write down the multiplicity
g1 (N1, s1 ) system 1
g2 (N2 , s2 ) system 2
so the multiplicity of the combined system is given by the following
N1 /2
g(N, s) =
∑
N1 /2
g1 (N1 , s1 )g2 (N2 , s2 ) =
s1 =−N 1 /2
∑
s1 =−N 1 /2
g1 (N1 , s1 )g2 (N2 , s − s1 )
We have removed S2 . These products have a maximum value which is to be determined for some value of
Sˆ1 , this is what is called the most probable configuartion (MPC). For large values of N we will show that
we need to keep only the most probable configuration of the sum. i.e
g1 (N1 , sˆ1 )g2 (N2 , s − sˆ1)
lets write this in terms of a Gaussian, we saw that
2
2 /N
2
g1 (N1 , s1 )g2 (N2 , s − s1 ) = g1 (N1 , 0)g2(N2 , 0)e−2s1 /N1 e−2(s−s1 )
Lets plot s1 versus g(N, s).The maximum value correspond to where these two curves intersect. This is
known as the most probable configuration.
s1
s2
s21
s22
=
=
=
=
sˆ1 + δ
sˆ2 − δ
sˆ1 2 + 2sˆ1 + δ2
sˆ2 2 − 2sˆ2 + δ2
Thus
2sˆ1 2 4sˆ1 δ 2sˆ1 δ 2sˆ1 2 4sˆ2 δ 2δ2
−
−
−
+
−
g1 (N1 , sˆ1 + δ)g2 (N2 , sˆ2 − δ) = g1 (0)g2(0) exp −
N1
N1
N1
N1
N2
N2
2δ2 2δ2
= (g1 g2 ) exp −
−
N1
N2
In the general case we know that
2
2 /N ]
2
g1 (N1 , s1 )g2 (N2 , s − s1 ) = g1 (N1 , 0)g2 (N2 , 0)e−[2s1 /N1 −2(s−s1 )
As we increase s1 this exponetial grows, thus we get a maximum value sˆ1 . What we need to do is figure
out this value, where we get the following result.
2δ2 2δ2
g1 (N1 , sˆ1 + δ)g2 (N2 , sˆ2 − δ) = (g1 g2 ) exp −
−
N1
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 that δ is
s
= 10−10
N1
δ = 1012
Thus
2s2
= 200
N
i.e
g1 g2
= e−400 ≈ 10−173
(g1 g2 )max
we can get a bit of insight in the following way, we can say that this configuration will never occur! The
system works its way through all possible values of g1 and g2 , so for this particular configuration to reach
the configuration g1 and g2 once, the system will be in (g1 g2 )max 10173 times. So if we suppose that any
spin flips every 10−12 seconds. Thus the number of configurations per second will be
1022
= 1034 /sec
10−12
so the time taken to get this deviation of
10173
δ
= 10−10 ∼
∼ 10139 sec ∼ 10132 yrs
N1
1034
where we know that the age of the universe is approximately 1018 seconds, thus we can conclude that this
configuartion will never occur. This is a very important result and tells us why we will only consider the
most probable configuartion MPC.
2.4.1 Find maximum value of g1 and g2
We know that g1 and g2 can be written in the form
2s2 2(s − s1)2
g1 g2 = g1 (0)g2(0) exp −
−
N1
N2
if we take the log of both sides we get
ln g1 g2 = ln[g1(0)g2 (0) −
2s2 2(s − s1 )2
−
N1
N2
if we take the derivative we find
∂
4s1 4(s − s1 )
s1
s − s1
s2
[ln g1 g2 ] = −
+
=0 ⇒
=
=
∂s1
N1
N2
N1
N2
N2
thus the fractional spin excess are equal, if we take the second derivative we find
4
4
∂2
[ln
g
g
]
=
−
−
<0
1
2
N1 N2
∂s21
21
so at
sˆ1
sˆ2
s
=
=
N1 N2 N
which is the fractional excess of the combined system, thus we have
sˆ2
s − sˆ1
s − sˆ1
sˆ1
=
=
=
N1 N2
N2
N − N1
which becomes
sˆ1
thus
1
1
+
N1 N − N1
=
s
N − N1
s
N1 − N1 + N1
=
sˆ1 =
N1 (N − N1 )
N − N1
which simplifies into
s
sˆ1
=
N1 N
using this result we find
sˆ1 2 sˆ2 2 sˆ1 sˆ1 sˆ2 sˆ2
s
s2
+
=
+
= (sˆ1 + sˆ2 ) =
N1 N2
N1
N2
N
N
thus we can conclude that
2 /N
(g1 g2 )max = g1 (0)g2(0)e−2s
2.5 Thermal Equilibrium: Entropy and Temperature
We have the expression for the multiplicity of the two systems given as
g(N, u) =
∑ g1 (N1, u1)g2(N2, u2)
u1 ≤u
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) =
∂g2
∂g1
g2 du1 +
g1 du2 = 0
∂u1
∂u2
where
du1 + du2 = 0
if we now take this expression and devide
1 ∂g1
1 ∂g2
=
g1 ∂u1 N1 g2 ∂u2 N2
this relates the derivative of the first system to the second system, hence
∂ ln g1
∂ ln g2
=
∂u1 N1
∂u2 N2
22
this is the condition for thermal equilibrium. This is where we will introduce entropy
σ(N, u) = ln g(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 ln g
as a final remark, Entropy is additive
σ(N, u) = ln[g1 (N1 , u1 )g2 (N2 , u2 )
= ln g1 (N1 , u1 ) + ln g2 (N2 , u2 )
= σ1 (N1 , u1 ) + σ2 (N2 , u2 )
We can define the temperature as
1
=
τ
thus in equilibrium
∂σ
∂u
N
τ1 = τ2
conventionally τ is written as
τ = kB T
we will relate τ to the Kelvin scale of absolute temperature.
2.5.1 Jusitification for keeping only the MPC
If we assume that
g = ∑ g1 (N1 , u1 )g2 (N2 , u2 ) = (g1 g2 )max
u1
we also assumed in our discussion of entropy that
σ = ln g(N, u) = ln(g1 g2 )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 expression
2δ2 2δ2
−
δ = s1 − sˆ1
g1 (N1 , s1 )g2 (N2 , s2 ) = (g1g2 )max exp −
N1
N2
lets suppose that we write
N1 = N2 =
N
2
we get
g(N, s) =
∑ g1(N/2, sˆ1 + δ)g2(N/2, sˆ1 − δ)
δ
= (g1 g2 )max
Z ∞
23
−∞
e−8δ
2 /N
dδ
so in our analysis we have just ignored the gaussian, the integral is
r Z
r
Z ∞
∞
2
N
Nπ
e−8δ/N dδ =
e−x dx =
8 −∞
8
−∞
hence
g(N, s) = (g1 g2 )max
r
Nπ
8
next we can use Sterling’s approximation
g(N, 0) = 2
N
2
πN
1/2
N→
N
2
we get the following
g(N/2, 0) =
2N/2
r
4
πN
!2 r
Nπ
4
= 2N
8
πN
r
Nπ
8
finally we need to form the entropy
4
σ(N, s) = ln g(N, s) = N ln 2 + ln
πN
1
Nπ
+ ln
2
8
we can rewrite this as follows
1
1
σ(N, s) = N ln 2 − ln N + ln N + ln(π/8) + ln(4/π)
2
2
1
= N ln 2 − ln N
2
if we take a big number N = 1022
N ln 2 ∼ 1022
ln N ∼ 2.3 log10 1022 ∼ 50
thus the only term we really care about is
σ(N, s) ≈ N ln 2
this term dominates all others for large N, thus we can conclude that only the most probable configuration contributes to the entropy and the temperature.
We have been talking about thermal equilibrium of two systems, the system was dominated by the
MPC. We have defined the Entropy and the temperature of the system. We will concentrate on is the laws
of thermodynamics
• Zeroth Law- two systems in thermal equilibrium with a third system then they are all in thermal
equilibrium
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 a
constraint is removed.
If we remember the two system multiplicity
g = ∑ g1 (u1 )g2 (u − u1 )
u
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))gi2(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(gi1 gi2 ) + ln
where
σi = ln(gi1 gi2 )
∑ g1 g2
we know that the final value of the entropy is
σ f ≥ σi
If we can imagine having two systems where τ1 > τ2 and that δu is removed from system and added to
system 2. What happens to the entropy when we do this
δσ
δσ
(−δu) +
(δu)
∆σ =
δu1 N1
δu2 N2
1
1
≥0
−
∆σ = δu
τ2 τ1
when τ1 > τ2 then δu > 0
• The Third Law of Thermodynamics- The entropy of a system approaches a constant value as the
temperature goes to zero.
This will follow from our statistical approach if temperature is an increasing function of energy
δT
>0
δu N
again we will have two systems where we have δu removed from system 2
δu1 = δu δu2 = −δu τ1 = τ2 = τ
again we have to go back to our law of increase of entropy, what happens to ∆σ1 and ∆σ2 , this are the
values at t = 0. We can do this by using a Taylor expansion
∂σ1
1 ∂2 σ1
σ1 = σ0 +
δu +
(δu)2 + ..
2
∂u1
2 ∂u1 u
u0
2 0
1 ∂ σ2
∂σ2
(−δu) +
σ2 = σ0 +
(−δu)2 + ..
2
∂u2
2
∂u
u0
2
u0
25
and
∆σ1
∆σ2
δu (δu)2 ∂
=
+
τ
2 ∂u
∂σ1
∂u1 u0
δu (δu)2 ∂ ∂σ2
= − +
τ
2 ∂u ∂u2 u0
and finally
∆σ = ∆σ1 + ∆σ2
is
∂
∂u
thus
∂σ
∂u
∂
=
∂u
1 ∂τ
1
=− 2
τ
τ ∂u
(δu)2
1 ∂τ1 1 ∂τ2
(δu)2 ∂τ1 ∂τ2
∆σ =
− 2
=− 2
<0
−
+
2
τ ∂u τ2 ∂u
2τ
∂u
∂u
the above function is evaluated at u0 , thus we have just shown that
∂τ
≥0
∂u
As τ is lowered then the energy decreases, thus at absolute zero τ = 0 then the system is in the lowest
energy 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 average
is simply
hxi = ∑ X (s)p(s)
s
which is a weighted average over all possible configurations
• Multiplicity function of combined system
g(N, u) =
∑ g1(N1, u1)g2(N2, u − u1)
u1 <u
for large N the sum was dominated by one term, and this term was called the most probable configuration MPC.
• From the multiplicity function we defined the entropy as
σ = ln(g(u)) S = kB σ
this was an extensive property, andsince we use the log then this made this an additive quantity.
26
• We then defined the temperature as
1
=
τ
∂σ
∂u
τ = kB T
N
• 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 that increases 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
1st
2nd
3rd
=
=
=
=
τ3 τ2 = τ3 τ1 = τ2
heat = energy
∆σ ≥ 0
lim σ = σ0
τ→0
2.7 Problems and Solutions
Problem # 1 Entropy and Temperature
Suppose g(U ) = CU 3N/2 , where C is a constant and N is the number of particles.
a) Show that
3
U = Nτ
2
Since we know
g(U ) = CU 3N/2
we can take the natural log of both sides
3N
3N/2
= ln(C) +
ln g(U ) = ln CU
ln(U )
2
Taking the partial derivative on both sides with respect to U yields
3N
∂
3N ∂
∂
ln(C) +
ln g(U ) =
ln(U ) =
ln(U )
∂U
∂U
2
2 ∂U
we know that the left hand term is simply the entropy
σ = ln g(U )
27
thus
3N 1
∂σ
=
∂U
2 U
thus
3N ∂U
2 ∂σ
U=
but we know that the temperature is defined as
1
=
τ
∂σ
∂U
N,V
thus we have just showed that
U=
b) Show that
3N
τ
2
∂2 σ
∂U 2
<0
N
show that it is negative. This form of g(U ) actually applies to an ideal gas.
We can simply take a double derivative of g(U )
∂
∂ 3N ∂
∂
ln g(U ) =
ln(U )
∂U ∂U
∂U
2 ∂U
this becomes
∂
∂2 σ
=
∂U 2 ∂U
3N 1
2 U
=−
3N
2U 2
since we know that N can only be positive and the energy can only be positive we have just shown
∂2 σ
∂U 2
<0
N
Problem # 2 Paramagnetism
Find the equilibrium value at temperature τ of the fractional magnetization
2hsi
M
=
Nm
N
of the system of N spins each of magnetic moment m in a magnetic field B. The spin excess is 2s. Take
the entropy as the logarithim of the multiplicity g(N, s) as given
σ(s) ≃ log g(N, 0) −
2s2
N
(2.1)
for |s| ≪ N. Hint : Show that in this approximation
σ(U ) = σ0 −
28
U2
2m2 B2 N
(2.2)
with σ0 = log g(N, 0). Further, show that 1/τ = −U /m2 B2 N, where U denotes hU i, the thermal average
energy.
Since we know that
σ0 = log g(N, 0)
thus
s=−
U = −2smB
U
2mB
thus Equation 1 becomes Equation 2, i.e
σ(U ) = σ0 −
U2
2m2 B2 N
if we take a differential on both sides with respect to U we find
∂σ(U )
U
=− 2 2
∂U
m B N
but we know that
U
1 ∂σ
=
=− 2 2
τ ∂U
m B N
so we find that the average thermal energy is given by
hU i = −
m2 B 2 N
τ
this is related to the spin excess and the magnetic moment by
hsi = −
hU i
mBN
=
2mB
2τ
we find that the equilibrium value at temperature τ of the fractional magnetization is given as
M
mB
=
Nm
τ
Problem # 3 Addition of entropy for two spin systems
Given two systems of N1 ≃ N2 = 1022 spins with multiplicity functions g1 (N1 , s1 ) and g2 (N2 , s−s1 ), the
product g1 g2 as a function of s1 is relatively sharply peaked at s1 = sˆ1 . For s1 = sˆ1 + 1012 , the product g1 g2
is reduced by 10−174 from its peak value. Use the Gaussian approximation to the multiplicity function;
the form
2
2
g1 (N1 , sˆ1 + δ)g2 (N2 , sˆ2 − δ) = (g1 g2 )max e
may be useful.
a) Compute g1 g2 /(g1g2 )max for s1 = sˆ1 + 1011 and s = 0.
29
2δ
− 2δ
N −N
1
2
(2.3)
For s1 = sˆ1 + δ we know that δ = 1011
δ = 1011
N1 ≃ N2 = 1022 s1 + s2 = s ⇒ s1 = −s2
We can see that Equation 3 can be written as
2
g1 (N1 , s1 )g2 (N2 , s − s1 )
= e−4δ /N1 ≈ 0.018
(g1 g2 )max
For s = 0 we know that δ = 0 thus we know that
g1 g2
= 1
(g1 g2 )max
b) For s = 1020 , by what factor must you multiply (g1 g2 )max to make it equal to ∑s1 g1 (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 , sˆ1 + δ)g2 (N2 , sˆ2 − δ) = (g1 g2 )max e
since
s1 = sˆ1 + δ
2
2
1
2
2δ
− 2δ
N −N
s2 = sˆ2 − δ
thus
2 2δ2
− 2δ
N −N
∑ g1(N1, s1)g2(N2, s − s1) = (g1g2 )max ∑ e
δ
s1
1
2
= (g1 g2 )max × Q
where Q is the factor we need to multiply (g1 g2 )max so that it is equal to ∑s1 g1 (N1 , s1 )g2 (N2 , s − s1 ). So
we can just turn this sum into an integral
Q=
Z ∞ 2δ2 2δ2 −N −N
e
−∞
1
2
dδ =
Z ∞ 4δ2 −N
e
1
dδ
−∞
this is simply a Gaussian integral with a solution
r
πN1
Q=
= 8.86 × 1010 ≈ 1011
4
Method 2
We know that
Q × (g1 g2 )max = ∑ g1 (N1 , s1 )g2 (N2 , s − s1 )
but we know that we can write the sum as
∑ g1 (N1, s1)g2(N2, s − s1) = g(N, s)
since we know that
N = N1 + N2 = 2N1
30
(2.4)
thus we find
2 /2N
1
= g(2N, 0)e−s
2 /2N
1
= g1 (N1 , 0)g2(N2 , 0)e−s
g(2N1 , s) = g(2N1 , 0)e−2s
2 /N
1
we also know that
(g1 g2 )max = g1 (N1 , 0)g2 (N2 , 0)e−2s
2 /N
1
thus Equation 4 can be written as
Q=
g(2N1 , 0)
∑ g1 (N1 , s1 )g2 (N2 , s − s1 )
=
(g1 g2 )max
g1 (N1 , 0)g2 (N2 , 0)
but we know that
g(N, 0) =
r
2 N
2
πN
and Equation 5 becomes
Q= q
q
2
2N1
2πN1 2
2
πN1
q
2 2N1
πN1 2
=
r
πN1
= 8.86 × 1010 ≈ 1011
4
c) How large is the fractional error in the entropy when you ignore this factor?
We know that the fractional error is defined as
∆σ σ′ − σ
=
σ
σ′
where
σ′ = ln(g1 g2 )max Q
σ = ln(g1 g2 )max
thus we find
∆σ
ln Q
ln Q
ln Q
=
=
=
σ
ln(g1 g2 )max Q ln Q + ln(g1 g2 )max ln Q + ln g1 (N1 , 0)g2(N2 , 0) − s2
N1
but we know that
2 /N
1
(g1 g2 )max = g1 (N1 , 0)g2 (N2 , 0)e−s
and we also know that
g1 (N1 , 0) =
r
2 N1
2
πN1
g1 (N1 , 0) =
r
2 N2
2
πN2
since we know that N1 = N2 = 1022 , plugging this into some calculating machine we find
∆σ
≈ 1.83 × 10−21
σ
this shows us that the extra factor does not affect the entropy very much.
31
(2.5)
Chapter 3
Boltzman Distribution & Helmholtz Free
Energy
• 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 temperature τ and also that the energy of the system is εl ≪ u0 . We must consider an entire ensemble of such
systems and then we must take the average of these. If we keep N fixed then we call this the cononical
ensemble. We must write down the multiplicity of the combinbed system
g(u0) = gR (u0 − εl )gs (εl )
what we are specifying with l is the label of a particular quantum state of a system. Thus there is only one
way for it to be in that state which means
gs (εl ) = 1
thus
g(u0 ) = gR (u0 − εl )
and since this total system is equally likely to be in any of of these states thus the probability is
pl = AgR (u0 − εl )
32
and since this is a probability
∑ pl = 1
l
again we have an approximation which implies that we will have to do a Taylor expansion, but we will
find that these multiplicity functions will begin to get out of hand when Taylor expanded. We will instead
Taylor expand the entropy
∂ ln[gR ]
+ ...
σ = ln[gR (u0 − εl )] ≈ ln[gR (u0 )] − εl
∂u
u0
εl
gR (u0 − εl )
=−
= ln
gR (u0 )
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 much
smaller than the energy in the resevoir
εl ≪ U0
where l is to denote the state of the system. What we are looking for is the ensemble average hX i
hX i = ∑ X (l)p(l)
l
we need to find the multiplicity function of the combined system
g(U0) = gR (U0 − εl )gs (l)
where
gs (l) = 1
by definition. Thus the probability we are looking for is proportional to g(U0), and to do this we must
Taylor expand to find
g(U0 − εl ) = Ae−εl /τ
where the probability is
p(l) ∝ e−εl /τ
which is known as the Boltzman factor.If we look at some simple model systems, i.e SHO
1
εn = n +
~ω
2
we can see that the higher the energy the lower the probability of finding this system in this state
pn+1 e−(n+3/2)~ω/τ
= −(n+1/2)~ω/τ = e−~ω/τ
pn
e
33
and we can see
pn+1
→0
pn
τ → 0
pn+1
→1
pn
τ →
and we also know that
∑ pl = ∑ Ae
−εl /τ
l
l
⇒A
∑e
−εl /τ
l
!−1
where
e−εl /τ
∑l e−εl /τ
where the denominator is known as the partition function
pl =
Z ≡ ∑ e−εl /τ
l
this is useful if you start with a microscopic model where we know how to calculate the microstate
energies. From these microstate energies we can use the partition function to get the thermodynamic
properties of this system. If we define the energy states then we can use the partition function as
Zn = ∑ gs (εn )e−εn /τ
n
where gs (εn ) is how many states have ε = εn . Lets find the average energy given the partition function
U ≡ hεi = ε̄
where
hεi = ∑ εl p(l) =
l
1
εl e−εl /τ
∑
Z l
where we often use β = 1/τ, using this we get
hεi = ∑ εl p(l) =
l
and the trick we can use
1
εl e−εl β
Z∑
l
∂ −εl β
e
= −εl e−εl β
∂β
and now we can write
∂Z
∂ ln Z
1
−
=−
hεi =
Z
∂β
∂β
to put this back in terms of τ we just simply need to use the chain rule
∂Z ∂Z ∂β ∂Z
1
− 2
=
=
∂τ
∂β ∂τ
∂β
τ
thus
∂ ln Z
1 2 ∂Z
τ
= τ2
Z ∂τ
∂τ
We will need to calculate what the scale of the energy fluctuations are, we will use a method that was
used in quantum mechanic.
hεi =
34
3.2 Energy Fluctuations
h(∆ε)2 i mean squared energy fluctuation
where
(ε − hεi)
and
h(∆ε)2i = hε2 − 2εhεi + hεi2i
= hε2i − 2hεihεi + hεi2
= hε2i − hεi2
we can use the same trick we have done before
hε2 i =
where the trick is
1
ε2l e−βεl
∑
Z l
∂2 −βεl
e
= ε2l e−βεl
∂β2
thus
1
hε i =
Z
2
∂2 Z
∂β2
∂
∂β
1 ∂Z
Z ∂β
1
=− 2
Z
∂Z
∂β
2
+
1 ∂2 Z
Z ∂β2
we can write this as
1
− 2
Z
to finish up
∂Z
∂β
2
1 ∂2 Z
1 ∂Z 2
+
+ hε2 i = −hεi2 + hε2 i
=−
2
Z ∂β
Z ∂β
∂
h(∆ε) i =
∂β
2
∂ ln Z
∂β
=
∂2 ln Z
∂β2
3.3 Paramagnetic System
We have N number of atoms in a volume where each atom has a spin of 1/2 and a magnetic moment m. We
apply a magnetic field H and we need to find hm(τ)i. The system consist of a single spin and the resevoir
has all the other spins with respect to τ. We will also neglect all the spin-spin interactions.
hm(τ)i =
1
ml e−βεl
∑
Z l
what are the states? there are only two states, 1 toward the field and one opposing the field
m↑ = +m
m↓ = −m
ε↑ = −mB spin up
ε↓ = mB spin down
and we also know that B = µH, now we find the partition function to be
Z = e−βε↑ + e−βε↓ = eβmB + e−βmB
35
so the expectation value is now
hm(τ)i =
meβmB − me−βmB
sinh(βmB)
=m
= m tanh(βmB)
βmB
−βmB
cos(βmB)
e
+e
thus
hm(τ)i = m tanh(βmB)
but we are looking for the moment for all N spins
βm
hM(τ)i = Nhm(τ)i = Nm tanh
τ
we can also find the suceptibility to be
dhMi Nm2
2 mB
χ=
=
sech
dB
τ
τ
if we look at the limits of the hyporbolic tangent function
tanh(x) =
thus for small x we find
lim tanh(x) =
x→0
and for large x we find
ex − e−x
ex + e−x
(1 + x) − (1 − x) 2x
=
=x
1+x+1−x
2
lim tanh(x) ≈ 1
x→∞
In the limit that mB/τ ≪ 1 or high temperature
Bm
hM(τ)i ≈ Nm
τ
and
χ=
Nm2
τ
=
Nm2 B
τ
Curie’s law
in the limit of low temperature we find
hM(τ)i = Nm
and
χ=0
another thing we are interested in is to find the average energy U = hεi
mB
hεi = −hM(τ)iB = −NmB tanh
τ
thus the heat capacity is defined as
CV =
∂U
∂T
kB
V
36
∂U
∂τ
=
∂U
∂T
where
∂U
mB
2 mB
− 2
= −NmBsech
∂τ
τ
τ
thus the heat capacity is
kB Nm2 B2
2 mB
sech
CV =
τ2
τ
lets look at what the heat capacity does in the extremes. At high temperatures
2
2
sech(x) = x
=
=1
−x
e +e
1+x+1−x
and
Nm2 B2
CV =
τ2
and at low temperatures
2
sech(x) = x = 2e−x
e
in this case the heat capacity is roughly
Nm2 B2 −mB/τ 2 4Nm2 B2 −2mB/τ
=
2e
e
CV =
τ2
τ2
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
1
χ→
τ − τC
where τC is known as the Curie temperature. This form tells us that when the temperature is lower than
the Curie temperature forces a negative susceptibility and we have a feromagnetic material and not paramagnetic.
At high temperatures the magnetic spins are all equally divided in up states and down states. At low
temperatures all of the spins are in the up (high energy) state and the magnetic energy is higher than the
thermal energy. If all the spins are aligned with he field it will lower the energy and if all the spins are
anti-aligned that raises the energy.
3.4 Calculating the average energy using the partition function
we can write the average energy as
hεi −
∂ ln Z
∂ ln Z
1 ∂Z
=−
= τ2
Z ∂β
∂β
∂Z
where we know that the partition function is given as
Z = emβB + e−mBβ = 2 cosh(mβB)
and thus
∂Z
= 2mB sinh(mβB)
∂β
so we have
1 ∂Z
2mB sinh(mβB)
=−
= −mB tanh(mβB)
Z ∂β
2 cosh(mβB)
which is the exact result we have already obtained.
hεi = −
37
3.4.1 Negative Temperature
We know that the energy is always an increasing function of temperature. We will need to consider a two
level quantum system. We can write the ratio of the number of particles in one energy state to the number
of particles in the other energy state
N2 e−E2 /τ
= −E /τ = e−(E2 −E1 )/τ > 1
N1 e 1
where E2 − E1 is assumed positive, for this to be true means that τ < 0. The negative temperature means
that we have a higher population in the high energy state than in the lower energy state. This is known as
a non-equilibrium system. This happens in lasers because we need more atoms in the high energy state
than in the lower energy state. To get laser action we must have all of the atoms in the upper state making
transitions to the lower energy state that. This is also know as population inversion. We also know that
negative temperatures require an upper bound on the energy, where this is given by NmB. This also means
that the multiplicity is bounded. This would not be true in the case of a gas, because as you raise the
temperature you also raise the energy. There is no negative temperatures for systems that have kinetic
energy or that the value of the multiplicity increases without bound. Negative temperatures do have more
energy than positive energies. If we have two systems, one with negative temperature and one with positive
temperature that are in thermal contact we find thet heat flows from the τ < 0 to τ >0.
3.5 Pressure
We define quantities such as σ and τ in terms of thermodynamic varaibles. But the pressure p must be
defined so as to be consistent with Newton’s laws.
Newton’s laws
If we have a cylinder with a gas at an initial position x and we expand the volume by an amount ∆x with a
constant area and a constant force pushing down on the gas. If the piston moves up by a distance ∆x then
we know that the work done on the gas is given by
∆W = −F∆x = −F
we also find this to be equal to
∆V
= −p∆V
A
∆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 state l. If we let V → V + ∆V (quasistatically), this means that the system
remains in thermal equilibrium. Imagine a box with distinct energy levels that are populated by the atoms
or 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 this
energy levels will not change. This is know as the adiabatical approximation where the state of the
system 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 as l, we can now do a Taylor expansion to find
∂εl
εl (V + ∆V ) = ε(V ) + dV
∂V σ
where the lower case sigma means that this process is at constant entropy. The ensemble average is given
by
∂U
U (V + ∆V ) = U (V ) + dV
∂V σ
= U (V ) − pdV
and thus
∂U
p=−
∂V σ
we need to know how to get pressure in terms of entropy. What we know is that we can write
σ = σ(U,V, N)
If we let
0 = dσ =
and now if we do
0=
where we know that
and we finally find
1
=
τ
∂σ
∂U
∂σ
∂U
where N is constant
∂σ
(dU )σ +
∂V
V,N
V,N
∂σ
∂U
∂U
∂V
p
=
τ
∂σ
∂V
(dV )σ
U,N
∂σ
+
∂V
σ
∂U
∂V
−p=
V,N
U,N
σ
U,N
3.6 Thermodynamic Identity
To simplify this we will write σ = σ(U,V ) and we will keep N fixed, thus
∂σ
∂σ
dU +
dV
dσ =
∂U V
∂V U
1
p
=
dU + dV
τ
τ
we can now write this as
τdσ = dU + pdV
thus we can write the change in the energy as
dU = τdσ − pdV
where τdσ 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 this
instead as
dU = d(τσ) − σdτ − pdV
where
d(U − τσ) = −σdτ − pdV
and
dF(τ,V ) = −σdτ − pdV
so the quantity
F(τ,V ) = U − τσ
this is known as Helmholtz free energy. If we take a derivative of this we get the following
∂F
∂F
dF(τ,V ) =
dτ +
dV
∂τ V
∂V τ
and hence
∂F
σ=−
∂τ
∂F
p=−
∂V
V
τ
where F is the appropriate energy for isothermal systems (fixed τ). If we look at the definition for the
pressure
∂F
∂(U − τσ)
∂U
∂σ
p=−
=−
=−
+τ
∂V τ
∂V
∂V τ
∂V τ
τ
where the first term is the energy part and the second term is the entropy part. The first tirm tends to
dominate for solids, wheras if you had a gas, this term would be really small. On the other hand the
second term dominates for a gas. There is one final thing we can get from this algebra, this is what is
called “the Maxwell’s relations”
3.8 The Maxwell Relations
From
If we differentiate
∂F
σ=−
∂τ
∂F
p=−
∂V
∂2 F
=−
∂V ∂τ
τ
∂σ
∂V
∂σ
∂V
thus we can see
V
τ
=
40
∂p
∂τ
∂p
∂τ
V
V
τ
∂2 F
=−
∂τ∂V
Why is the Helmholtz free energy interesting? We know that the free energy is a minimum for a system in
thermal equilibrium with a resevoir.
F = U − τσ
imagine a resevoir where
dFs = dUs − τσs
fixed temperature
we also have the following
1
=
τ
∂σ
∂U
dUs = τσs constant volume
V
if we substitute this into the above equation we find
dFs = 0
is this a maximum or a minimum? Lets consider the total energy of the system
U = UR +Us
σ = σR + σs = σR (U −Us ) + σs (Us)
if we assume Us ≪ UR then we can simply do a Taylor expansion
∂σR
+ σs (Us )
σ = σR −Us
∂UR V
we can recall that
1
=
τ
∂σR
∂UR
V
putting this in gives
1
σ = σR (U ) − (Us − σs τ)
τ
we can see that
Fs
τ
this is because that we are assuming that σR (U ) is a fixed quantity. We know that in thermal equilibrium
that the entropy is a maximum, therefor, if we consider a small deviation from σ
σ = σR −
∂σ = ∂σR −
∂Fs
∂Fs
=−
τ
τ
thus for τ > 0 then if σ is a maximum means that Fs is a minimum. On the other hand the energy of the
system corresponds to a minimum.
Example: Paramagetism
Lets find the Helmholtz free energy F and the average magnetization M of N spins at τ ≫ mB (high
temperature). We can write the free energy as
F(s, τ, B) = U (s, τ, B) − σ(s, τ, B)τ
41
where we know that
U (s, τ, B) = −2smB
and also
2 /N
g(N, s) = g(N, 0)e−2s
this is the Gaussian approximation in the high temperature limit. Lets put all this together
F(s, τ, B) = −2smB − τ ln g(N, 0) +
2s2 τ
N
finally, in thermal equilibrium we can take the derivative
4sτ
∂F
= −2mB +
=0
∂s τ,B
N
thus
2s =
mBN
τ
where
m2 BN
M = 2sm =
τ
this 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
σ=−
∂τ
F = U − τσ
thus
and we also know
∂F
F =U +τ
∂τ
∂
τ
∂τ
2
and we find
V
V
∂F
F
= −F + τ
τ V
∂τ V
∂
F =U +τ
∂τ
F
+F
τ V
2
and finally
∂
U = −τ
∂τ
2
but recall that
U = τ2
F
τ V
∂ ln Z
∂τ
and thus we find
F
= − ln Z
τ
42
this is a nice way to relate the microscopic properties Z to the macroscopic properties F. We can also write
this expression as
Z = e−F/τ
if we remember the probability
e−εl /τ
= e(F−εl )/τ
Z
thus we are able to calculate the Boltzman factor if we know the free energy of the system without ever
knowing what the partition function is.
p(εl ) =
3.10 One Particle in a Box
For simplicity we assume that our box is a cube, with sides l. There is one particle with mass M. We all
know how to write down the Schrodinger Equation and we can find the energies as
~2 π 2 2
(nx + n2y + n2z )
εm =
2m L
n2 = n2x + n2y + n2z n2x + n2y + n2z ≥ 1
we can now write the partition function
Z1 = ∑ ∑ ∑ e
−
~ 2 π2
(n2 +n2y +n2z )
2mL2 τ x
nx ny nz
since we know that the temperature is much greater than te energy spacing allows us to turn this sum into
an integral. This is know as the classic limit approximation. If we let
~2 π 2
2mL2 τ
α2 =
thus we can write
Z1 =
Z ∞
0
dnx
Z ∞
0
dny
but this can also be written as
Z1 =
Z
∞
Z ∞
0
−α2 n2x
e
0
2 (n2 +n2 +n2 )
x
y
z
dnz e−α
dnx
the solution is
3
π3/2 23/2 M 3/2 L3 τ3/2
= V nQ
8
~3 π 3
where nQ is known as the quantum concentration and defined as
Z1 =
nQ =
Mτ
2π~2
43
3/2
3.11 Average energy
U =τ
now
ln Z1 =
thus
or
2
∂ ln Z1
∂τ
V
3
ln τ + τ independent terms
2
∂ ln Z1
31
=
∂τ
2τ
V
U = τ2
3
3
3
= τ = kB T
2τ 2
2
and the specific heat is given by
3
CV = kB
2
this allows us to measure the mean square velocity
1
3
Mhv2 i = kB T
2
2
thus
3kB T
M
We want to know what the meaning of the concentration is? Recall the Debroigle wavelength
hv2 i =
λD =
h
h
h
≈
≈
p M(3kB T /M)1/2 (3Mτ)1/2
and
λ3D ≈
1
h3
≈
3/2
nQ
(3Mτ)
thus
nQ ∼
1
λ3D
this means that the concentration corresponds to one particle in a volume of λ3D . This allows us to understand degeneracy in gases. If 2 particles are separated by a distance much greater then λD then their wave
functions do not overlap and they are in the “classical” regime, and the results do not depend whether they
are bosons or fermions. On the other hand if they are separated by less than the Debriogle wavelength
then their wave functions overlap and they are in the “quantum” regime. Now the results depend crucially
on whether they are bosons or fermions.
3.11.1 Example: N particles in a box
Imagine that we have a number of particles N2 at standard pressure and temperature STP. we know that
n≈
6 × 1023 molecules/mole
≈ 3 × 1019 molecules/cm3
22.4 × 103 cm3 /mole
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 × 300
2π(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
3
U= τ
2
and that “each” “square” term in momentum and position coordinates has a mean energy τ/2 (kB T /2) in
the classical limit. For an ideal gas, the hamiltonian is given by
H=
p2y
p2
p2x
+
+ z
2m 2m 2m
and
3
U= τ
2
Lets imagine that we only have one dimension. Thus the Hamiltonian is given by
H=
and we have an energy given by
p2 1 2
+ kx
2m 2
U =τ
Proof
Lets imagine a one dimensional gas with a Hamiltonian
H=
p2
= Bp2i
2m
which is just a general expression for a quadratic term. The energy is given by
∂ ∞ −βεi
−βεi d p
− ∂β
d pi
−∞ e
i
−∞ εi e
ε̄i = R ∞ −βε
= R ∞ −βε
id p
id p
i
i
0 e
−∞ e
R∞
this is simplified as
∂
ε̄i = − ln
∂β
R
Z
∞
−∞
45
εi e
−βεi
d pi
and now
Z ∞
−∞
if we let
d pi =
Z ∞
−∞
2
εi e−βBpi d pi
y = β1/2 pi
we get
Z ∞
−∞
but we know that
εi e
−βεi
"
1
ln p
β
Z ∞
−∞
εi e
−βεi
−By2
εi e
1
d pi = p
β
Z ∞
−∞
2
εi e−By dy
#
1
dy = − ln β + terms independent of β
2
thus the average energy is
∂
1
1
τ
ε̄i = −
− ln β =
=
∂β
2
2β 2
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 of R and if we were to measure the noise across a VN ,
so as a function of time we would get something in which its average is zero, but the mean square value
is not zero. We could then measure the current noise by connecting a wire at the end of the circuit. We
will once again see a signal that fluctuates. This provides an absolute limitation on how accurately you
can measure something. We will apply the equipartition function to see how this works. We will calculate
the power spectrum which is also known as the spectral density. This is usually given as Sν ( f ) but in our
case it is convenient to write it as SV (ω) the following relationships are
SV (ω)dω = SV ( f )d f
where
df
dω
we want to look at a narrow window that has a 1 Hz bandwidth, thus this is the mean square voltage in a
1 Hz bandwidth, as you will see, as we scan the frequency the value of S will vary. If we can imagine a
circuit with a resistor, a conductor and a capacitor. The quality factor for this circuit is given by
Sv (ω) = SV ( f )
Q=
ω0 L
R
ω20 =
1
LC
we can also write Q as
R
ω0
∆ω =
∆ω
L
Now using the equipartition we can see that there will be two square terms, thus we get the following
Q=
and
1
1
ChVN2 i = kb T
2
2
hVN2 i =
1 2
1
LhIN i = kB T
2
2
hIN2 i =
46
kB T
C
kB T
L
all frequencies add up to contribute to the square terms. The task is to go from the mean square term to
the spectral density. So Lets write down themean square value for the current source
hIN2 i =
Z ∞
SV (ω)dω
kB T
=
L
|Z(ω)|2
0
this takes a voltage squared and devide by the impedence squared. We all remember that we can write
Z(ω) = R + iωL +
and we also rember that
1
iωC
1
= ω20 L
C
thus we find that
Z(ω) = R +
but
2
2
|Z(ω)| = L
this allows us to write
hIN2 i =
iL
(ω − ω20 )2
ω
R2 (ω2 − ω20 )2
+
L2
ω2
Z ∞
SV (ω)
L2
0
R2
∆ω = 2
L
2
dω
∆ω2 +
(ω2 −ω20 )2
ω2
this function is sharply peaked about ω0 for high Q. If we take the limit we find
SV (ω) → SV (ω0 )
if we look at
(ω2 − ω20 )2 ((ω − ω0 )(ω + ω0 ))2
=
ω2
ω2
if we let ω → ω0 we find
(ω2 − ω20 )2
= 4(ω − ω0 )2
ω2
given this we get the following result
SV (ω0 )
kB T
=
L
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
where we all know that
SV (ω0 )
4L2
Z ∞
Z ∞
dx
0
thus we find
0
dx
x2 + a2
x2 + a2
=
π
a
kB T
SV (ω0 )π
SV (ω0 )
= 2
=
π
L
4L (R/2L)
2LR
47
thus we find that the spectral
2
SV (ω0 ) = kB T R
π
this answer does not depend on either L or C, thus this is true for all values of L and C. This is true for all
values of ω. This is avery good example of what is called “white noise”. Thus we have the final result
2
SV (ω) = kB T R
π
SV ( f ) = 4kb T R
this is the famous results called Nyquist noise or Johnson noise. The current noise is given by
SI ( f ) =
4kB T
R
The Nyquist noise sets a fundamental limit on the smallest voltage or current that one can measure. For
example if we have some experiment that can be modeled with a resistor connected to an amplifier then
the smallest value of the voltage will always be limited by the Nyquist noise. What we will see is that
the magnitude scales linear with the absolute temperature. Thus to make more accurate measurements we
must decrease the tempearture. This enables us to measure lower voltages or anything else we might want
to measure. If we lower the temperature enough then we will reach the quantum limit of the noise, which
will take over when h f ≥ kB T . As a final remark because of the linear dependence with temperature than
the Nyquist noise is used as an absolute thermometer.
3.14 Summary
The Boltzman factor
e−εl /τ
Z
is the probability using the Boltzman factor. We then introduced the partition function
p(εl ) =
Z = ∑ e−εs /τ
s
we then intruduced the concept of pressure, the pressure is given by
∂σ
∂U
=τ
p=−
∂V σ
∂V U
where we have the identity given by
dU = τdσ − pdV
this allows us to write the Helmholtz free energy
F = U − στ
this is only for isolated thermal systems. We also defined the entropy as
dF
σ=−
dτ V
48
and
dF
p=−
dV
the Maxwell relations are given by
where the free energy is
∂σ
∂V
τ
=
dp
dτ
τ
V
F = −τ ln Z
and the probability is given by
p(εl ) = e(F−εl )/τ
we then went to a particle in a box, where
Z1 = nQV
nQ =
Mτ
2π~2
3/2
where the energy of one particle is given by
3
U= τ
2
we then move on to the Equipartition theorem
U=
τ
2
per quadratic term. Finally we talked about the Nyquist noise
SV ( f ) = 4kB T R
SI ( f ) =
4kB T
R
3.15 Problems and Solutions
Problem # 1
A dilute solution of macromolecules at temperature T is placed in a centrifuge rotating with angular
velocity ω. The mass of each molecule is m. The equivalent centrifugal force on each particle in the
rotating frame of reference is mω2 r, where r is the radial distance from the axis of rotation.
Find how the relative density of molecules ρ(r) varies with r.
The probability that a system will be in a specific quantum state s of energy εs is proportional to the
Boltzman factor. The ratio of the probability that the system is an a quantum state 1 at energy ε1 to the
probability that the system is in quatum state 2 at energy ε2 is just the ratio of the two multiplicities
P(ε1 ) gR (U0 − ε1 )
=
P(ε2 ) gR (U0 − ε2 )
(3.1)
the multiplicity is the number of states having the same value s. This is a direct consequence of what is
called the fundemental assumption. If the resevoir is very large, than the multiplicities are very large. We
can write Equation 1 in terms of the entropy of the resevoir
σ = ln gR
49
gR = eσ
thus Equation 1 now becomes
we can define
P(ε1 ) e(σR (U0 −ε1 ))
=
P(ε2 ) e(σR (U0 −ε2 ))
(3.2)
∆σR = σR (U0 − ε1 ) − σR (U0 − ε2 )
(3.3)
this allows us to write Equation 2 as
P(ε1 )
= e∆σR
P(ε2 )
since we know that U0 ≫ ε allows us to Taylor expand the entropy (Equation 3). The Taylor expansion is
defined as
1 2 d2 f
df
+ a
+ ...
f (x0 + a) = f (x0 ) + a
dx x=x0 2!
dx2 x=x0
thus we find
σR (U0 − ε1 ) =
=
σR (U0 − ε2 ) =
=
thus we find
σR (U0 ) − ε1 (∂σR /∂U )V,N + ...
σR (U0 ) − ε1 /τ
σR (U0 ) − ε2 (∂σR /∂U )V,N + ...
σR (U0 ) − ε2 /τ
∆σR = −(ε1 − ε2 )/τ
and we finally find that
P(ε1 )
= e−(ε1 −ε2 )/τ
P(ε2 )
thus the ratio of the number of particles in a particular quantu state 1 to the number in a particular state 2
is given by the ratios of their probabilities
N1 P(ε1)
=
= e−(ε1 −ε2 )/τ
N2 P(ε2)
we know that the density is defined as
ρ=
thus we know that
ρ1 =
m
N
V
m
N1
V
ρ2 =
m
N2
V
and the relative density is given by
ρ̄ =
ρ1 N1
=
= e−(ε1 −ε2 )/τ
ρ2 N2
which is simply the Boltzman factor. Since we also know that the centrifugal force is defined as
Fc = mω2 r
and the energy is given by
ε=−
Z
1
F · dr = − ω2 r2
2
50
thus we know that
ρ̄ =
ρ(r)
= eε1 /τ
ρ(0)
thus we find the relative density to be given as
1
2 r2
ρ̄ = e 2τ mω
a plot is given as
Problem # 2
Show that for a system in thermal contact with a resevoir σ = ln Z +U /τ. This is a very useful result,
as you will see in (b) below.
Consider a crystalline solid containing N atoms whose nuclei have spin one, so hat each nucleus can
be in one of the three states ms = 1, 0, −1. We assume that the electric charge distribution in the nucleus
is ellipsoidal, so that the energy of a nucleus depends on its spin orientation with respect to the internal
electric field of the crystal. Thus, the energy is ε for ms = 1 or -1, and zero for ms = 0.
a) Find an expression for the nuclear contribution to the internal energy, U .
First we need to show that
σ = ln Z +
U
τ
we can do this by knowing that the free energy is defined
F = U − στ
F = −τ ln Z
putting these two expressio equal to each other yields
−τ ln Z = U − στ
51
which reduces to
σ = ln Z +
U
τ
Next we need to find the average energy of the system (ensemble average) this is given by
U = hεs i = ε̄
where
hεi = ∑ εl p(l) =
l
1
εl e−εl /τ
∑
Z l
where we often use β = 1/τ, using this we get
hεi = ∑ εl p(l) =
l
and the trick we can use
and now we can write
1
εl e−εl β
Z∑
l
∂ −εl β
e
= −εl e−εl β
∂β
∂Z
∂ ln Z
1
−
=−
hεi =
Z
∂β
∂β
to put this back in terms of τ we just simply need to use the chain rule
1
∂Z ∂Z ∂β ∂Z
− 2
=
=
∂τ
∂β ∂τ
∂β
τ
thus
hεi =
1 2 ∂Z
∂ ln Z
τ
= τ2
Z ∂τ
∂τ
Thus the energy is given by
U = hεi = τ2
∂ ln Z
∂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 function Z, of a single nucleus, and hence the partition function Z for N
nuclei. Hence, find the entropy of the solid.
We know that the partition function is defined as
Z = ∑ e−εs /τ
s
and we know the ms = ±1 εs = ε and for ms = 0 εs = 0. Thus we find thet the partition function for this
system is given by
ZN = (1 + 2e−ε/τ )N
and the entropy of the solid is given by
σ = N ln Z +
52
U
τ
using Equation 4 and the partition function we find the entropy to be given as
σ = N ln(1 + 2e−ε/τ ) + τ
now we just need to find
∂ ln(1 + 2e−ε/τ )
∂τ
∂ ln(ZN ) N ∂Z
=
∂τ
Z ∂τ
and we find
∂
∂
(1 + 2e−ε/τ ) = 2 (e−ε/τ )
∂τ
∂τ
letting
u=−
ε
∂u
= 2
∂τ τ
ε
τ
thus we find
∂ ln(1 + 2e−ε/τ ) 2Nε e−ε/τ
=
∂τ
τ 1 + 2e−ε/τ
thus we find the total entropy to be given as
τ
σ = N ln(1 + 2e−ε/τ ) +
2Nε e−ε/τ
τ 1 + 2e−ε/τ
a plot is given by
Where we plot τ/ε instead of ε/τ.
c) By directly counting the number of accessible states, calculate the entropy as τ → 0 and τ → ∞. Show
that your expression in (b) reduces to these values.
Since we know that the probability goes as
p=
1
g
g=
1
p
thus the multiplicity for 1 particle goes as
53
and the multiplicity for N particles is given by
N
1
g(N) =
p
and as τ → ∞ we know all the states are equally probable thus
g(N) = 3N
and thus at this limit the entropy is
p=
1
3
σ = ln g = N ln 3
at the other extreme when τ → 0 we know that all particles will all be in the ground state and thus the
probability is
p = 1 g = 1N
and the entropy is
σ = ln g = N ln 1 = 0
now to show this with our previous solutions we just need to take the limits of our function to find
In the limit that τ goes to 0
lim ln(1 + 2e−ε/τ ) +
τ→0
2ε e−ε/τ
≈0
τ 1 + 2e−ε/τ
this makes sense because we know if the temperature goes to 0 then the entropy goes 0.
In the limit that τ → ∞ we find
lim ln(1 + 2e−ε/τ ) +
τ→∞
2ε e−ε/τ
≈ ln 3
τ 1 + 2e−ε/τ
which also makes sense, as the temperature goes to infinity the entropy converges to a constant number.
d) Calculate the heat capacity CV of the crystal, and find its temperature dependence for high τ. Make a
sketch of CV 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
we need to find
e−ε/τ
∂ ln Z
= 2Nε
∂τ
1 + 2e−ε/τ
∂
∂U
= 2Nε
∂τ
∂τ
54
e−ε/τ
1 + 2e−ε/τ
!
Using the quotient rule
∂
∂x
g f ′ − f ′g
f
=
g
g2
if we say
f = e−ε/τ
f′ =
ε −ε/τ
e
τ2
and
g = 1 + 2e−ε/τ
thus
g′ =
2ε −ε/τ
e
τ2
"
#
∂U
(1 + 2e−ε/τ )(ε/τ2 )e−ε/τ − e−ε/τ (2ε/τ2)e−ε/τ
= 2Nε
∂τ
(2e−ε/τ + 1)2
this can be simplified by factoring out a term
"
#
2Nε2 −ε/τ 1 + 2e−ε/τ − 2e−ε/τ
e−ε/τ
2Nε2
∂U
= 2 e
=
∂τ
τ
τ2 (1 + 2e−ε/τ )2
(2e−ε/τ + 1)2
thus the heat capacity for this system is given by
CV =
2Nε2
e−ε/τ
τ2 (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=
we can write the heat capacity as
CV = 2Nx2
ε
τ
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 of τ of a system with two states, one at energy 0
and one at energy ε.
Since we know that the partition function gives
Z = 1 + e−ε/τ
and the free energy is defined as
F = −τ ln Z
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 = ln Z +
but we know the energy is defined as
U = τ2
U
τ
∂ ln Z
∂τ
56
thus
σ = ln Z + τ
where
∂ ln Z ε
=
τ
∂τ
τ
∂ ln Z
∂τ
e−ε/τ
1 + e−ε/τ
!
thus the entropy is defined as
σ = ln(1 + e
−ε/τ
ε
)+
τ
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 the magnetization M and the susceptibility
χ ≡ dM/dB as a function of temperature and magnetic field. The result for the magnetization is
M = nm tanh(mB/τ), as derived in (46) by another method. Here n is the particle concentration.
We know that the average magnetic moment is given as
hm(τ)i =
1
ml e−βεl
∑
Z l
57
what are the states? there are only two states, 1 toward the field and one opposing the field
ε↑ = −mB spin up
ε↓ = mB spin down
m↑ = +m
m↓ = −m
and we also know that B = µH, now we find the partition function to be
Z = e−βε↑ + e−βε↓ = eβmB + e−βmB
so the expectation value is now
hm(τ)i =
meβmB − me−βmB
sinh(βmB)
=
m
= m tanh(βmB)
cos(βmB)
eβmB + e−βmB
thus
hm(τ)i = m tanh(βmB)
but we are looking for the moment for all N spins
Bm
hM(τ)i = Nhm(τ)i = Nm tanh
τ
but we know that
Nhm(τ)i
Bm
M(τ) =
= nm tanh
V
τ
n=
N
V
thus
Bm
M(τ) = nm tanh
τ
we can also find the suceptibility to be
dhMi nm2
2 mB
=
sech
χ=
dB
τ
τ
thus
nm2
2 mB
sech
χ=
τ
τ
b) Find the free energy and express the result as a function only of τ and the parameter x ≡ M/nm.
We know that the free energy is given as
F = −τ ln Z
and we know that the partition function is given as
Z = eβmB + e−βmB
thus the free energy is given as
βmB
F = −τ ln(e
−βmB
+e
mB
) = −τ ln 2 cosh
τ
58
and to put it in terms of x ≡ M/Nm we simply do
M
mB
x=
= tanh
Nm
τ
thus
mB
= tanh−1 (x)
τ
we also know that
and finally we find that
cosh2 (x) − sinh2 (x) = 1 cosh(x) = q
1
1 − tanh2 (x)
2
)
F = −τ ln( √
1 − x2
c) Show that the susceptibility is χ = nm2 /τ in the limit mB ≪ τ.
we know that for large x we find
lim tanh(x) ≈ 1
x→∞
In the limit that mB/τ ≪ 1 or high temperature
Bm
hM(τ)i ≈ nm
τ
and
χ=
nm2
τ
we plot M/nm vs mB/τ
59
=
nm2 B
τ
Problem # 6
N molecules of N2 are condensed onto a surface of area L × L. The molecules do not interact with each
other, so each molecule is completely free to move translationally in 2 dimensions (along the surface) and
to vibrate, with vibrational frequency ν = 1012 Hz and non-degenerate energy levels ε = s~ω with s
being any positive integer (ignoring zero point energy). Assume the N2 are distinguishable and that L is
large so that
~2 π 2
≪ ~ω
2M L
the following relationship will be useful
∞
a
∑ aqn = 1 − q
n=0
for |q| < 1
this is known as a geometric series.
a) Write down an exact expression for the energy U as a function of temperature τ. You do not need to
derive this expression but please justify either in words or from simple mathematical expressions
where it comes from.
We know that the translational energy given by a 1-D ideal gas is given by
1
U = Nτ
2
but since we have a 2-D system (2 degrees of freedom) we find the total translational energy to be
Utran = Nτ
This comes from solving for the energy levels and wavefunctions for particles in an L × L box. For N
distinguishable particles, the partition function is
ZN =
"
∞
∑ (e−~ω/τ)s
s=0
#N
and using the expression for the geometric series we find
1
ZN =
1 − e−~ω/τ
N
we know that the energy is given as
Uvib = τ2
∂ ln Z
(1 − e−~ω/τ ) −~ω/τ ~ω
N~ω
= τ2 N
= ~ω/τ
e
2
−~ω
2
∂τ
τ
(1 − e
)
e
−1
thus the total energy of this system is given by
Utot = Nτ +
60
N~ω
e~ω/τ − 1
b) Calculate the exact specific heat CV as a function of temperature τ and plot this showing units and
values on the axes.
We know that the specific heat is defined as
N~ω
~ω
∂U
= N + ~ω/τ
e~ω/τ 2
CV =
2
∂τ V
τ
(e
− 1)
thus the specific heat is given by
"
~ω
CV = N 1 +
τ
2
e~ω/τ
(e~ω/τ − 1)2
#
Problem # 7
Consider N 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 energy ε2 = ε and is triply degenerate. The system
is in thermal contact with a resevoir at temperature τ.
a) Find the partition function for the system.
We know that the partition function of this system is given by
Z = ∑ e−εs /τ = 1 + 3e−ε/τ = 1 + 3e−βε
s
and for N particles we find
b) Find U, F, and σ as a function of τ.
h
iN
−βε
ZN = 1 + 3e
We know that the energy is defined as
U = τ2
1 ∂Z
∂ ln Z
=−
∂τ
Z ∂β
we know what Z is now we need to find
∂Z
= N[1 + 3e−βε ]N−1 [−3εe−βε ]
∂β
which can simplify into
U=
3Nε
eε/τ + 3
To find the free energy we know that
F = −τ ln ZN = −τN[1 + 3e−ε/τ ]
and the entropy is given as
U
1
3Nε
−ε/τ
σ = ln ZN + = N ln[1 + 3e
]+
τ
τ eε/τ + 3
61
c) Find the specific heat CV as a function of τ and calculate its low and high τ limits.
We know that the heat capacity is given as
CV =
where we can see that
∂U
∂τ
V
ε 2
eε/τ
3Nε
∂U
ε/τ ε
= ε/τ
=
3N
e
∂τ
τ2
τ (eε/τ + 3)2
(e + 3)2
thus the heat capacity is given as
CV = 3N
for high temperature limit τ ≫ ε we can see that
ε 2
τ
eε/τ
(eε/τ + 3)2
ε
eε/τ → 1 + ≈ 1
τ
thus
3N ε 2
CV =
16 τ
for the low temperature limit we can see that τ ≪ ε gives
eε/τ ≫ 3
thus
CV = 3N
ε 2
τ
e−ε/τ
Problem # 8
Consider again the system described in Problem 7. For each particle, the ground state is a spin singlet
with s = 0 and ε = 0 and the first excited state is a spin triplet with s = 1. With the applied magnetic field
B = 0, the three s = 1 states have equal energy ε as described above. (The energysplitting ε in B = 0 is due
to effects other than magnetiuc field, specifically what are called crystal field splittings). In a magnetic
field, the s = 0 state still has energy 0 (ms = 0), but the s = 1 states split into three states (ms = ±1, 0) ,
with energies εs = ε + 2µB B, ε, ε − 2µBB for ms = −1, −, +1 respectively where µB is the Bohr magneton.
a) Calculate the partition function as a function of B and τ.
We know that the partition function for this system is given by
ZN = ∑ e−εs /τ = [1 + e−(ε+2µB B)/τ + e−(ε−2µB B)/τ + e−ε/τ ]N
s
b) Calculate the expectation value of the moment M of the system as a function of field B (at fixed
temperature τ). Hint M = Nhms i
62
Since we know that the expectation of the moment is given as
1 ∞
1
M = Nhms i = N ∑ ms e−εs τ = N [e−(ε−2µB B)/τ − e−(ε+2µB B)/τ ]
Z s=0
Z
this becomes
M=N
e−(ε−2µB B)/τ − e−(ε+2µB B)/τ Ne−ε/τ 2µB B/τ
Ne−ε/τ 2 sinh(2µB B/τ)
=
[e
− e−2µB B/τ ] =
ZN
ZN
ZN
Problem # 9
Consider a system that has 5 possible states, with two energy levels. The lowest energy level is doubly
degenerate (two states) with energy ε1 = 0. The next energy level has threefold degeneracy (three states)
with energy ε2 . It is in thermal contact with a resevior at temperature τ.
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 energy ε = 0 is given by
2
e−ε/τ
=
p(ε = 0) =
Z
Z
where the partition function is given by
Z = ∑ e−εs /τ = 2 + 3e−ε2 /τ
s
thus the probability is given as
p(0) =
2
2 + 3e−ε2 /τ
b) Calculate the energy U and the entropy σ (as a fucntion of τ).
We know that the energy is given by
U =−
where
1 ∂Z
Z ∂β
∂Z
= −3ε2 e−βε2
∂β
thus the energy is given as
3ε2 e−ε2 /τ
U=
2 + 3e−ε2 /τ
And we know that the entropy is defined as
"
#
−ε2 /τ
e
3ε
U
2
σ = ln Z + = ln[2 + 3e−ε2 /τ ] +
τ
τ 2 + 3e−ε2 /τ
thus we find
"
#
−ε2 /τ
3ε
e
2
σ = ln[2 + 3e−ε2 /τ ] +
τ 2 + 3e−ε2 /τ
63
c) Calculate the energy and the entropy as τ goes to 0 and τ goes to infinity. Briefly discuss/explain your
answers.
We can see that as τ goes to 0 we find
e−ε2 /τ → 0
thus
U → 0 σ → ln 2
as τ goes to 0, the system goes into the ground state which have ε = 0 ⇒ U → 0. As τ → ∞
e−ε/τ → 1 −
thus we find
U→
ε
τ
3ε2 (1 − τε )
3ε2 (1 − ε/τ)
ε =
2 + 3(1 − 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
∂σ
∂σ
dU +
dV
dσ(U,V ) =
∂U V
∂V U
and the thermodynamic definition
dU = τdσ − pdV
F = U − τσ
calculate dF. From this, calculate p and σ in terms of derivatives of F (show your calculations of
the derivatives explicitely, i.e where they come from, don’t just state the result) and from this, prove
the Maxwell relation.
We know that
dσ(U,V ) =
∂σ
∂U
∂σ
dU +
∂V
V
we also know that
thus
dV =
U
dU p
+ dV
τ
τ
dU = τdσ − pdV
and
where
F = U − τσ
dF = dU − τdσ − σdτ = τdσ − pdV − τdσ − σdτ = −pdV − σdτ
∂F
∂V
τ
= −p
64
∂F
∂τ
V
= −σ
if we take the cross-derivatives
which yields
∂
(p)
∂τ
∂p
∂2 F
=−
∂τ∂V
∂τ V
∂
(σ)
∂τ
∂σ
∂2 F
=−
∂v∂τ
∂V τ
the order of the secon derivatives do not matter and thus we find
∂σ
∂p
=
∂τ V
∂V τ
b) There is one important variable which is held constant in the above 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 calculating F, σ and p
and their derivatives for the ideal monatomic gas. Start from Z for the ideal monatomic gas, which
you do not need to re-derive
1
Z = (nQV )N nQ =
N!
Mτ
2π~2
3/2
We know that the free energy is given by
F = −τ ln Z = −τ ln
1
(nQV )N = −τ[− ln N! + N ln nQ + N lnV ]
N!
we also know that the entropy is defined as
∂F
∂ ln nQ
1 ∂nQ
3 nQ 3
σ=−
= τN
= τN
= τN
= N
∂τ V,N
∂τ
nq ∂τ
2 τ
2
thus
and for the pressure we find
3
σ = [− ln N! + N ln nQ + N lnV ] + N
2
∂F
p=−
∂V
and we can finally show that
and
∂σ
∂V
τ,N
τ,N
= τN
∂p
∂τ
=N
∂ lnV
N
=
∂V
V
V,N
65
∂ lnV
τN
=
∂V
V
=
N
V
Chapter 4
Thermal Radiation and The Planck
Distribution
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 filled with EM radiation, if it is in thermal equilibrium then the modes only depend on the temperature. We need to calculate this which gives rise to the
Planck distribution. The first thing we need to do is find the density of modes. The second thing is to find
the mean oocupancy or mean energy of each mode using the Boltzman distribution. What are modes? You
get resonances with the length of the cavity or of the string is equal to an integer of its wavelength
L1 =
λ
3λ
L2 = λ L3 =
2
2
or a general expression is given by
2
λ=n ,
L
n = 1, 2, 3, ...
where
πc
2πc
=n
λ
L
to do this calculation we need to regard each mode as being equivalent to the simpe harmonic oscillator of
energy levels at that frequency. The energy of the SHO is given by
ω = 2π f =
ε = n~ω n = 1, 2, 3, ...
where n is the number of photons in that energy states.
66
4.2 Density of modes
Lets consider a cavity with sides L, lets have the waves propgate in the z direction. We have the electric
field and the magnetic field that are orthogonal to each other. There are also two different polarization that
are independent. Lets consider (for example) Ex , next we write down the wave equation
2
∂2
∂2
1 ∂2 Ex
∂
2
+
+
E
=
∇ Ex =
x
∂x2 ∂y2 ∂z2
c2 ∂t 2
but we know that
n πx n πy n πz y
x
z
sin
sin
L
L
L
writing this in this form already places boundary conditions. Where they are that Ez vanishes on all the
walls. Now we must substitute Ex into the wave equation. We get the following
Ex (x, y, z,t) = Ex0 sin(ωt) sin
−
π2 2
ω2
2
2
(n
+
n
+
n
)
=
−
y
z
L2 x
c2
this gives us a relationship between all of then, ω, and the cavity. Thus
n2 =
L
L2 2
ω n= ω
2
2
π c
πc
n≥0
we need to remember that this is in three dimensions. We now want to calculate the density of nodes for
this box.
D(ω)dω = number of modes between ω, ω + dω
we now need to cosider an octant of a sphere, in fact we want to consider a shell that has a thickness dn
and a radius that we will call n, this shell contains a certain number of nodes. We need to find
1
D(ω)dω = 2 × 4πn2 dn = πn2 dn
8
the 2 comes about from the two independent modes of polirization, then we need to consider that the
surface area is 1/8 times the surface of the sphere and the dn, thus we have
D(ω) = πn2
but we know what this is
thus
dn
dω
dn
L
=
dω πc
3
L2 2 L
L
D(ω) = π 2 2 ω
ω2
=π
π c
πc
πc
and we find
D(ω) =
V 2
ω
π2 c3
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 energy kB T per mode because of the two quadratic term
in E 2 and B2 , this immideatley gives us a result for the energy in the black body spectrum
V 2
ω kB T
π2 c3
u(ω) = D(ω)kB T =
if we integrate over the entire spectrum to get the energy we will get the following
uT (ω) =
Z ∞
0
D(ω)kB T dω =
V
kB T
π2 c3
Z ∞
0
ω2 dω
we can see that this integral diverges. This cannot be true. It was know that the spectrum does not go to
infinity. The classical theory worked well at low frequencies, but not at high frequencies. Then came Max
Planck who introduced the high frequency limit
uT (ω) ∼ ω2 e−~ω/kB T
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 mode in our box with frequency ω, we want
to know what the average energy at a temperature T . 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=
∑ e−nε/τ
n=0
where n is the number of photons in the mode and the energy is quantized. Thus ε = n~ω, this is really the
difference between the classical theory and quantum theory. Lets find the mean occupancy of that mode
hni =
−nε/τ
∑∞
n=0 ne
−nε/τ
∑∞
n=0 e
if we write
y = e−ε/τ
thus
hni =
there is a simple trick to do this
hni = y
∑ nyn
∑ yn
∞
d
ln ∑ yn
dy n=0
these two expressions are equivalent. Now we know that
∞
1
∑ yn = 1 − y
n=0
68
and so we are left with
hni = −y
thus
hni =
and thus we get the result
d
ln(1 − y)
dy
y
1
=
1 − y 1/y − 1
hni =
1
e~ω/τ − 1
this is known as the Planck distribution. There are two points to make, This function takes care of the
ultraviolet catastrophe
1. Occupancy tends to zero as ω goes to zero
2. This whole theory dependend on the fact that we quantized the energy. Thus EM radiation can be
represented by photons of discrete energy. The results implies that the energy is in fact quantized in
units of ~ω (“photons”). This was the introduction in Quantum theory.
We can now write the energy density as
u(ω)dω = hni~ωD(ω)dω
where ~ω is th energy of the mode and hni is the average number of photons with some energy. Thus
u(ω)dω =
V ~ω3 dω
π2 c3 (e~ω/τ − 1)
we can write this more suggestively as
x=
this gives
~ω
τ
V ~ τ 4 x2 dx
u(ω)dω = 2 3
π c ~
ex − 1
thus we can see that the energy goes to the fourth power of the temperature. It will also give us a very useful
scaling argument. The derivative of this function gives us the turning point of the energy distribution. We
find the maximum to be given by
2 x
~ωmax
d
xmax =
= 2.82
x
dx e − 1)
τ
this gives us the following scaling law This results gives us what is known as Wien’s law
~ωmax ~ωmax
=
τ1
τ2
this enables one to measure temperatures from the black body spectrum, e.g stars and pyrometers. Lets
look 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
V ~ω3
V ω2 τ
=
π2 c3 (~ω/τ)
π2 c3
first we can see that ~ drops out and also this is precisely the Rayleight-Jeans result.
u(ω) =
• The Quantum limit ~ω/τ ≫ 1, energy levels are very discrete
u(ω) =
V ~ω3 −~ω/τ
e
π2 c3
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 energy
in the cavity is given by
Z
uτ =
∞
u(ω)dω
0
it is convinient to take the formed introduced previoulsy
Z
V ~ τ 4 ∞ x3
uτ = 2 3
dx
x
π c ~
0 e −1
π2V τ4
∝ τ4
=
3
3
15~ c
where we have said
Z ∞
π4
x3
dx
=
x
15
0 e −1
thus we find the total energy of the cavity is given by
uτ =
π2V τ4
15~3 c3
this result is independent of the nature of the cavity or of materials placed inside of it. One othet thing that
we can work out is the entropy.
4.5 Entropy
We know that
τdσ = du + pdV
thus, at constant volume we have
dσ =
where
uτ =
π2V τ4
15~3 c3
and so we find
dσ =
duτ
τ
duτ =
4π2V τ3
dτ
15~3 c3
4π2V τ2
dτ
15~3 c3
if we now integrate both sides we find
σ=
4 π2V τ3
45 ~3 c3
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
1
N(ω) = n(ω)c
4
we can compare this with the kinetic theory of gases. With this in mind it is simle to write the power
emmited for the blackbody. Where the power emmited per unit area in the frequency range of ω + δω is
given by
1
Pl (ω)dω = n(ω)c(~ω)dω
4
now we know that
uT (ω)
energy per unit volume
n(ω)~ω =
V
thus the total energy is given by the following
total emmited power =
Z ω
0
1
Pl (ω)dω = c
4
where
utot
Z ω
uT (ω)dω
0
V
=
1 utot
4 V
π2V τ4
=
15~3 c3
thus we find
π2 τ4
= σSB T 4
60~3 c2
this expressions defines the Stephan-Boltzman constant
Ptot =
π2 kB4
W
σB =
= 5.67 × 10−8 2 4
3
2
60~ c
m K
this quantity is know very precisely. We will now spend a few minutes talking about absorption and
emmision 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 property
in terms of an absorption coefficient.
a(ω) = 1
aborptivity
we also have what is called a grey body
a(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 emmisivity e(ω), which is simply the fraction of electromagnetic radiation compared to
that 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 have
vibration. There are two energy levels. This means that we absorb and emit strongly at some frequency ω.
This is know as the principle of detailed balance. There are lots of good example of this, i.e a silver tea
pot should be polished to keep your tea warm for a long time. This is because if the tea pot is very shiny
than 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 that
UV 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 trapped
inside the green house. The glass absorbs and emmits strongly in the infrared, hence the green house
warms up. There is also “Global Warming”, this is due to the gases and clouds in the atmosphere. On
clear nights it gets colder than if it is cloudy. The clouds act as an absorbing and emmiting medium for
infrared radiation.
4.8 Phonos: Debye Model
If we think of latice waves then we are really thinking about sound waves, thus they propogate at the
velocity of sound vs , there are three possible independent modes. The first is a longitudinal wave, back
and 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 to
figure 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 of
possible modes. On the other hand, sound waves in a solid, the number of modes is bounded. Each
atom has three degrees of freedom, thus the maximum number of modes for each atom there are 3 modes
and for N atoms there are 3N modes. If we consider a single transverse wave, it takes discrete points to
identify that wave. If we try to draw the frequency with only 2 points, then this has no meaning. There is
a minimum wavelength. We can recall that for photons (2 modes)
V ω2
D(ω) = 2 3
π c
but for phonons (3 modes)
D(ω) =
3V ω2
2π2 v3s
there is an additional twist. This is cut off at ωD , which is the Debye frequency (highest frequency where
we can have vibrations). Thus
3N =
Z ωD
0
3V
D(ω)dω = 2 3
2π vs
72
Z ωD
0
ω2 dω =
V ω3D
2π2 v3s
thus the Debye frequency is given by
1/3
2 3 N
ωD = 6π vs
V
this results is independent of the size of the solid. The velocity of sound is
v3s
V ω3D
= 2
6π N
if we plug this into the previous equation we find
D(ω) =
3V ω2 6π2 N
ω2
=
9N
2π2 V ω3D
ω3D
thus this scales as ∝ Nω2 where ω2 as for photons. The second step is to figure out the population of these
modes.
4.9 Total Energy
The total energy at temperature T is given by
utot (T ) =
Z ωD
D(ω)~ωhnidω
0
where hni is the average number of phonos in a mode, ~ω is the phonon energy, and D(ω) is the density
of modes. Thus we find
Z
9N~ ωD ω3 dω
utot (T ) = 3
ωD 0 e~ω/kB T − 1
so we can write this integral in a dimensionless form
x=
~ωD kB TD θD
~ω
, xD =
=
=
kB T
kB T
kB T
T
where θD is the “Debye temperature”
θD =
~ωD
kB
ω=
xkB T
~
dω =
kB T
dx
~
thus we will get the following
9N(kB T )4
u(T ) =
~3 ω3D
where
Z xD
0
x3
dx
ex − 1
x3
dx
x
0 e −1
is dimension-less, this can only be solved numerically or by lookin at the extremes (limits).
Z xD
73
4.9.1 Low temperature
xD ≫ 1 or kB T ≪ ~ωD = kB θD , we can replace xD by infinity in the upper limit, and hence
Z ∞
π4
x3
dx
=
ex − 1
15
0
thus in this limit
u(T ) =
9N(kB T )4 π4
~3 ω3D 15
u(T ) =
3π4 N(kB T )4
5(~ωD )3
thus
this is the total energy of a solid at low temperatures. It is more useful to talk about the heat capacity
12π4 NkB kB T 3
∂U
=
CV =
∂T V
5
~ωD
thus
3
12Nπ4
T
CV =
kB
5
θD
this is called the Debye T 3 law. All non-magnetic insulator obey this law. But metals do not obey this
because they have electronic contribution.
4.9.2 High Temperature
xD ≪ 1 or kB T ≫ ~ωD or T ≫ θD , this is a simple case
Z xD
0
thus we can expand ex as 1+x
Z xD
x3
dx
ex − 1
x2 dx =
0
x3D
3
and we find the total energy to be
u(T ) =
9N(kB T )4 x3D
~3 ω3D 3
which yields
u(T ) = 3NkB T
wher 3N is the total number of atoms in the crystal. We have N atoms that act as 1D harminic oscillator,
this is in fact a 3D problem. This is the equipartition theorem applied to N 3 dimensional simple harmonic
oscilator. We can easily derive the heat capacity
CV = 3NkB
this is a very important results, this is known in chemistry as the Dulong and Petit law which is
≈ 25 joule deg−1 mole−1
74
we can also plot CV vs the temperature we find, at low temperature we have the famous T 3 law and at high
temperature we find that this is independent of temperature. θD is a measured quantity, a few examples
are given as
Indium
108K
Copper
343K
Diamond 1500K
there is a clear temperature dependence and hardness of the material. Hard materials are “stiff” therefore
they have a high value of spring constant, thus high ωD ⇒ high θD . We should expect that there should
also be a contribution from the electric contribution to the heat capacity. The metals have free electrons
that apperently do not contribute significantly to CV at room temperature.
4.10 Fluctuations in Phonon Energy for Temperatures ≪ θD in T 3
Limit
We have a resevoir and a system, we want to see how the fluctuations scale with temperature. We remember
that
∂u
∂u
2 ∂u
2
2
=τ
= (kB T )
= kB T 2CV
h(∆u) i = −
∂β
∂τ
kB ∂T
we need to recall the expressio for the energy in this limit
u(T ) =
3π4 (kB T )4 N 3π4 kB T 4 N
=
5(~ωD )3
5
θ3D
we know that the energy is in the form of
u = AT 4
thus the heat capacity is
CV =
and now we can write
∂u
∂T
= 4AT 3
20kB T θ3D
h(∆u)2 i kB T 2 4AT 3 4kB T
20 1 θ3D
=
=
=
=
=
u2
uAT 4
u
3π4 kB T 4 N 3π4 N T 3
20
3π4
θD
T
3
1
N
if we take a small thin film with l = 100µm w = 10µm, and h = 0.1µm which gives us a volume of
V = 10−16 m3
where the typical density is ≈ 1028 m3 , thus we hav e approximately N ≈ 1012 atoms, and also
θD ≈ 300 K
T = 0.01 K
if we put all these number in we find
1
h(∆u)2 i
≈
u2
15
300
0.01
3
× 10−12 ≈ 2
thus we have large fluctuations in the energy, in fact, fluctuations results break down because the energy
of 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) =
V ω2
π2C3
we then found the famous Planck distribution
hni =
1
e~ω/kB T
−1
and the energy density is given by
u(ω) = hni~ωD(ω) =
and the total energy is given by
uτ =
V ~ω3
π2 c3 (~ω/τ)
π2V τ4
15~3 c3
We also found the total power to be given as
P = σSB T 4
where
σSB =
π2 kB4
= 5.678 × 10−8 Wm−2 K−4
3
2
60~ c
we also derived Kirchoffs laws
a(ω) = e(ω)
we also discussed phonos
3
12π4
T
CV =
NkB
5
θD
and
CV = 3NkB
T ≪ θD
T ≫ θD
4.12 Problems and Solutions
Problem # 1
A wire of a radius r0 is coincident with the axis of a metal cylinder of radius R and length L. The
wire is maintained at a positive potential V with respect to the cylinder. The whole system is at some high
temperature T . As a result, electrons emitted from the hot metals form a dilute gas filling the cylindrical
container and in equilibrium with it. The density of these electrons is so low that their mutual electrostatic
interactions can be neglected.
76
a) Use Gauss’s theorem to obtain an expression for the electrostatic field which exists at points at a radial
distance r from the wire (r0 < r < R). The cylinder of length L may be assumed to be very long so
that end effects are negligable.
We know that Gauss’s law is given by
Z
where we know that
Z
E · da =
qenc
ε0
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
Z R
Z
1
qenc
R
qenc
dr =
ln
V = E · dl =
2πLε0 r0 r
2πLε0
r0
thus
qenc =
2πLε0
V
ln(R/r0 )
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 variable density which fills the entire space between
the wire and cylinder. Using the result of part (a), find the dependence of the electric charge density
(relative density) on the radial distance r.
We know that the relative density is given by the Boltzman factor, since we can assume that there are N1
and N2 particles having energy ε1 and ε2 respectively then we can write the relative density as
ρ̄ =
ρ(r1)
= e−(ε(r1 )−ε(r2 ))/τ
ρ(r2)
77
we have proved this result elsewhere. Thus we need to find what the energy is at distance r1 and r2 . We
know that the energy due to some charge and potential is given by
ε(r) =
thus we find that
p2
− qV (r)
2m
∆ε = q[V (r2 ) −V (r1 )]
if we let
V (r) = C
Z r
1
r
dr = C ln
r
r0
r0
thus
r1
V (r1 ) = C ln
r0
r2
V (r2 ) = C ln
r0
this allows us to find thet the change in energy is given by
r1
r2
− ln
∆ε = Cq ln
r0
r0
where we know that C is given from Equation 1 as
C=
V
ln(R/r0 )
thus we find the relative density to be given as
ρ(r1)
−Cq
= e−∆ε/τ = e
ρ̄ =
ρ(r2)
h i
r
r
ln r2 −ln r1 /τ
0
0
if we choose appropriate radius r1 = r0 and r2 = r2 we find the the relative density is given by
ρ(r0 )
− V q
= e ln(R/r0)
ρ̄ =
ρ(r2 )
h i
r
ln r2 /τ
0
Problem # 2 Zipper Problem
A zipper has N links; each link has a state in which it is closed with energy 0 and a state in which it
is open with energy ε. We require, however, that the zipper can only unzip from the left end, and that the
link number s 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 by s = (1, 2, 3, ..., s − 1) and we also know that the
total energy is given by εs = nε, thus we can write the partition function as
N
Z=
∑ e−εs /τ =
s=0
N
∑ e−nε/τ =
n=0
78
N
∑ (e−ε/τ)n
n=0
this resembles a geometric series of the form
k
∑ arn = a
n=0
1 − 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 the
unwinding 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
hNi =
∑ np(l) =
n=0
−nε/τ
∑N
n=0 ne
Z
if we expand the numerator to the first few terms we find
hNi = (e−ε/τ + 2e−2ε/τ + 3e−3ε/τ )(1 − e−ε/τ )
if we let
x = e−ε/τ
we find that the average number of open links is given by
hNi = (x + 2x2 + 3x3 )(1 − x)
Problem # 3 Elasticity of Polymers
The hermodynamic identity for a one-dimensional system is
τdσ = dU − f dl
when f is the external force exherted on the line and dl is the extension of the line. By analogy with
∂σ
p=τ
∂V U
we form
∂σ
f = −τ
∂l
U
the direction of the force is opposite to the conventional of the pressure, this is due to the fact that (∂σ/∂l)
is always negative.
We consider a polymeric chain of N links each of length ρ, with each link equally likely to be directed
to 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!
( 21 N + s)!( 21 N − s)!
we know that the multiplicity is defined as
g(N, s) =
N!
( 21 N + s)!( 21 N − s)!
we can also see that
l = 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 the total multiplicity of the system), we simply
get
2N!
g(N, −s) + g(N, s) = 1
( 2 N + s)!( 21 N − 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 given by
2 /N
g(N, s) ≈ g(N, 0)e−2s
and since we know that
s=
l
2ρ
we 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
N
in 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)] −
80
l2
2Nρ2
c) Show that the force at extension l is
f=
lτ
Nρ2
The force is proportional to the temperature. The force arises because the polymer wants to curl up: the
entropy is higher in a random coil than in an uncoiled configuration. Warming a rubber band makes it
contract; 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
∂σ
f = −τ
∂l
we simply need to differentiate the entropy as a functio of length, which yields
∂σ
l
=− 2
∂l
Nρ
Substituting this into the force yields
f=
lτ
Nρ2
Problem # 4 Earth’s Steady State Temperature
The surface temperature of the Sun is T0 = 5500 K; its radius is Rsun = 7 × 1010 cm while the radius of
the Earth is Rearth = 6.37 ×108 cm. The mean distance between the Sun and the Earth is r = 1.5 ×1013 cm.
In the first approximation one can assume that both the Sun and the Earth absorb all electromagnetic
radiation incident upon them. The Earth has reached a steady state so that its mean temperature T does
not change in time in spite of the fact that the Earth constantly absorbs and emits radiation.
a) Find an approximate expression for the temperature T of the Earth in terms of the astronomical
paramters mentioned above.
We know that the power recieved at the Earth from the sun is given by
P(earth) = Fsun Aearth =
Lsun 2
Lsun 2
πR
=
R
earth
4πr2
4r2 earth
where r is the distance from the Sun to the Earth r = 1 AU. We also know that the luminosity of an object
is given by
4
Lsun = 4πR2sun σTsun
thus the power the Earth receives from the sun is given by
P(earth) =
4 R2
πR2sun σTsun
earth
2
r
we know that the power emmited by the Earth is given
4
P(earth) = 4πR2earth σTearth
setting these two expressions together we find that the temperature of the Earth is given by
4 R2
πR2sun σTsun
4
earth
= 4πR2earth σTearth
2
r
81
thus
Tearth = Tsun
r
Rsun
2r
b) Calculate this temperature T numerically.
Since we know that Rsun = 7 × 108 m and r = 1.5 × 1011 m and Tsun = 5500 K we find the temperature of
the Earth to be
s
(7 × 108 )
Tearth = 5500 K
2(1.5 × 1011)
this is simply
Tearth = 265.67 K
Problem # 5 Number of Thermal Photons
Show that the number of photons ∑hsn i in equilibrium at temperature τ in a cavity of volume V is
N = 2.404π−2V (τ/~c)3
where we know that the entropy is given by
σ=
4π2V
45
τ 3
~c
whence σ/N ≃ 3.602. It is believed that the total number of photons in the universe is 108 larger than the
total number of nucleons (protons, neutrons). Because both entropies are of the order of the respective
number of particles, from
5
σ = N ln(nQ /n) +
2
the photons provide the dominent contribution to the entropy of the universe, although the particles dominate the total energy. We believe that the entropy of the photons is essentially constant, so that the entropy
of the universe is approximately constant with time.
We know that
hsn i =
thus we can see that
1
e~ωn /τ − 1
N = ∑hsn i = ∑
n
n
1
e~ωn /τ − 1
where the sum is over the triplet of integers nx , ny , nz . Positive integers alone will describe all independent
modes. We replace the sum over nx , ny, nz by an integral over the volume element dnx , dny, dnz in the
space of the mode indices. That is we set
N=
Z ∞
0
n2 dn
dΩ
e~ωn /τ − 1
82
where dΩ is the solid angle given by
dΩ =
Z 2π
dφ
0
Z π
0
thus we find
N = 4π
Z ∞
4π
N=
8
Z ∞
sin θdθ = 4π
n2 dn
0 e~ωn /τ − 1
to make our lives easier, lets just integrate over one octant, this gives us
0
since we know that
nπc 2
ωn =
n =
L
we are now able to write
N=
π
2
Z ∞
0
if we introduce the unitless quantity
x=
π~cn
Lτ
n=
n2 dn
e~ωn /τ − 1
Lωn
πc
2
dn =
L
dω
πc
n2 dn
enπc~/Lτ − 1
Lτ
x
π~c
dn =
Lτ
dx
~πc
using these identities we are now able to write
3 Z ∞ 2
τ 3 Z ∞ x2
Lτ
1
x
−2
N=
dx = π V
dx
x
~c π2 0 ex − 1
~c
0 e −1
where we have looked up the solution to
Z ∞
0
thus we find
x2
dx ≈ 2.404
ex − 1
N = 2.404π−2V
τ 3
~c
Problem # 6 Average Temperature of the Interior of the Sun
a) Estimate by a dimensional argument or otherwise the order of magnitude of the gravitational selfenergy of the Sun, with Msun = 2 × 1033 g and Rsun = 7 × 1010 cm. The gravitational constant G
is 6.6 × 10−8 dyne cm2 g−2 . The self-energy will be negative referred to atoms at rest at infinite
seperation.
The energy due to a gravitational field produced my some mass M is given by
U =−
GM 2
r
given the quantities for M, G and R we find
U =−
GM 2
≈ −3.81 × 1041 Joules
r
83
b) Assume that the total thermal kinetic energy of the atoms in the Sun is equal to − 12 times the gravitational energy
1
K =− U
Virial Theorem
2
this is a result of the Virial Theorem in mechanics. Estimate the 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 range in central temperature for different
stars, excluding only those composed of degenarate matter for which the law of perfect gases does
not hold (white dwarfs) and those which have excessively small average densities (giants and super
giants), is between 1.5 and 3.0 × 107 degrees.”
Since we know that the total therma kinetic energy is given by
1
K = NkB T = − U = 1.9 × 1041 joules
2
thus we find that the temperature will be given as
T =−
U
1.9 × 1041 joules
=
2NkB
NkB
since we know the number of particles and what the constant kB we find that the temperature is
T=
1.9 × 1041 joules
K ≈ 7.24 × 106 K
1 × 1057 · 1.38 × 10−23 joules
Problem # 7 Angular Distribution of Radiant Energy Flux
a) Show that the spectral density of the radiant energy flux that arrives in the solid anle dΩ is cuω cos θ ·
dΩ/4π, where θ is the angle the normal to the unit area makes with the incident ray, and uω is the
energy denisty per unit frequency range.
84
Since we know that the differential energy dE is given by
dE = cdtuωdΩdA
where we also know that the incident beam will not contribute anything to the flux if the angle of
incidence is 90 degrees or π/2. Thus I know that my differential area dA is really given by dA = cos θdA,
this will account for any angle dependence on the incidence of the beam. Thus we find
dE = cdtuω dΩ cos θdA
or this can also be written as
dE
= cuω cos θdΩdA
dt
we must also normalize this over all solid angle 4π ster-radians, thus we find
dE
dΩ
= cuω cos θdA
dt
4π
we also know that the flux is defined as
F(ω) =
dE
dtdA
thus we find that the flux density is given as
F(ω) = cuω cos θ
85
dΩ
4π
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 =
cuω
4π
Z 2π
0
this is given as
cuω
2
Ftot =
we know that the integral
dφ
Z π/2
Z π/2
0
Z π/2
0
sin θ cos θdθ
sin θ cos θdθ
1
2
0
thus we find the total flux integrated over the solid angle is given by
sin θ cos θdθ =
1
F = cuω
4
Problem # 8 Free energy of a photon gas
a) Show that the partition function of a photon gas is given by
Z = ∏[1 − e−~ωn /τ ]−1
n
where the product is over the modes n.
We know that the partition function for a photon in a mode n is given by
∞
Z = ∑ e−~ωn /τ
n
This sum is of the form ∑ xn with x ≡ exp(−~πc/τL). Because x is smaller than 1, the infinite series may
be summed and has the value
1
= [1 − e−~ωn /τ ]−1
Z=
−~ω
n /τ
1−e
since we know that for many photons in a gas the partition function can be written as
Zn = Z1 Z2 Z3 ...Zn
since this are distinguishable photons in a given energy state n. Thus the partition can be expressed as a
product of the form
Z = ∏[1 − e−~ωn /τ ]−1
n
b) The Helmholtz free energy is found directly as
F = τ ∑ ln[1 − e−~ωn /τ ]
n
86
Transform the sum to an integral; integrate by parts to find
F =−
π2V τ4
45~3 c3
turning this sum into an integral yields
1
F = 2 × τ4π
8
Z ∞
0
n2 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 integrating over one octant and the 4π comes from integrating over the entire solid angle of a sphere. We
also know that
πnc
ωn =
L
thus the free energy (in integral form) is
F = τπ
Z ∞
0
n2 [1 − e−~nπc/Lτ ]dn
letting
Lτ
~nπc
dn =
dx
x=
Lτ
~nπc
2
n =
Lτ
~πc
2
x2
yields
τ
F= 2
π
Lτ
~c
3 Z
∞
0
x2 ln[1 − e−x ]dx
integration by parts yields
τ
F= 2
π
Lτ
~c
3 x3
1 − e−x
3
∞
0
1
−
3
Z ∞
0
x3
dx
1 − e−x
\right]since we know that
x3
ln[1 − e−x ]∞
0 →0
3
Z ∞
0
π4
x3
dx
=
1 − e−x
15
we find that the free energy is
F =−
π2V τ4
45~3 c3
Problem # 9 Heat capacity of liquid 4 He at low temperatures
The velocity of longitudinal sound waves in liquid 4 He at temperatures below 0.6 K is
2.383 × 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 for 4 He we must begin with the total number of elastic modes
equaling N
Z
4π nmax 2
n dn = N
8 0
we find that
1/3
6N
π 3
nD = N nD =
6
π
the thermal energy of the phonon is
U = ∑hεn i = ∑hsn i~ωn = ∑
~ωn
~ω
e n /τ − 1
for this problem we find the thermal energy to be given as
4π
U=
8
if we let
Z nD
~ωn
n2 dn
~ω
/τ
n
e
−1
0
ωD =
πvnD
L
we find the energy to be given as
U=
if we let
~πvnD π
L 2
Z nD
0
π~nD v π~v
=
xD =
Lτ
Lτ
n2D
e~πnD v/Lτ − 1
6N
π
where θ is called the Debye temperature expressed as
1/3
=
dn
θ kB θ
=
T
τ
1/3
~v
2N
θ=
6π
kB
V
since we know that
ρ
Mtot N Mtot 1
=
=
m1 V
V m1 m1
where m1 is the atomic mass unit of a helium atom. Since we know
N=
~
kB
m1
v
ρ
=
=
=
=
=
1.05 × 10−34 Js
1.38 × 10−23 J/K
4 × 1.66 × 10−24 g
2.383 × 104 cm s−1
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 value
CV = 0.0204 × T 3 , in J g−1 K−1 . The T 3 dependence of the experimental value suggest that phonos
are the most important excitations in liquid 4 He below 0.6 K. Note that the experimental value has
been expressed per gram of liquid.
The energy in the low temperature limit for this case is given as
U=
π4 Nτ4
5(kB θ)3
we can see that the heat capacity is given as
4π4 Nτ3
4 4
T3
∂U
=
CV =
=
π
k
N
B
∂τ V 5(kB θ)3 5
θ3
to find the heat capacity per gram we use
4π4 kB N
CV
=
×T3
3
g
5θ mass
where
1
N
=
mass m1
where m1 is the atomic mass unit for helium, thus we find
4π4 kB
CV
= 3 × T 3 = 0.0211 J g−1 K−3
g
5θ m1
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 a
permeable material where particles can go through it. We have
N1 , u1 and N2 , u2
where
u = u1 + u2 N = N1 + N2
these are both constant. We will proceed by writing down the most probable configuration MPC occurs
when
g1 (N1 , u1 )g2 (N2 , u2 )
is a maximum. If we diferentiate this, we know that
∂g1
∂g2
∂g2
∂g1
dN1 +
du1 g2 +
dN2 +
du2 g1
d(g1 , g2 ) =
∂N1
∂u1
∂N2
∂u2
we will also now say
du1 = −du2 dN1 = −dN2
if we now divided by g1 , g2 we find
1 ∂g1
1 ∂g2
1 ∂g1
1 ∂g2
−
dN1 +
−
du1 = 0
g1 ∂N1
g2 ∂N2
g2 ∂u1
g2 ∂u2
90
but now we note the following
1
g1
∂g1
∂u1
=
∂ ln g1
∂N1
=
∂σ1
∂N1
if we now this for all these terms we get the following
∂ ln g2
∂σ2
1 ∂g2
=
=
g2 ∂u2
∂N2
∂N2
thus we find
∂σ1
∂N1
∂σ2
−
∂N2
dN1 +
∂σ1
∂u1
∂σ2
−
∂u2
du1 = 0
now we know that dN1 and du1 are independent. Hence each term in the brackets is independent of 0.
1
∂σ2
1
∂σ1
=
⇒ =
∂u1
∂u2
τ1 τ2
but we find a new term
∂σ1
∂N1
∂σ2
∂N2
=
µ
− =
τ
∂σ
∂N
thus we can now define the chemical potential as
u
this is the chemical potential. The chemical potential has dimensions of energy. We know that particles
will diffuse from a higher µ until µ1 = µ2 at thermal equilibrium. We can remove δN particles from 2 and
add to 1. Thus
δN1 = −δN2 = δN > 0
and
δσ = δσ1 + δσ2 =
if initially
µ
∂σ1
∂σ2
µ2 1
δN
δN −
∂N = − +
∂N1
∂N2
τ
τ
µ2 > µ1
thus the entropy would increase and δσ is positive and thus we are moving to thermal equilibrium. As a
consequence particles will flow from higher µ to lower µ 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
∂σ
∂σ
du +
dV +
dN
∂u
∂V
∂N
91
thus
1
p
µ
dσ = du + dV − dN
τ
τ
τ
we ca now simply write this
du = τdσ − pdV + µdN
this is now our thermodynamic 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
∂u
µ=
∂N σ,V
lets recall the Helmholtz free energy
F = u − τσ
if we differentiate we find
dF = du − τdσ − σdτ
= τdσ − pdV + µdN − τdσ − σdτ
dF = −σdτ − pdV + µdN
thus we find
dF = −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 consider µ as the
increase in the internal energy u when 1 particle is added to the system at constant σ and V . It is also the
increase in the Helmholts free energy to the system at constant τ and V . You can have both internal and
external 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 initially µ1, N1 , τ1 and the second has µ2 , 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 charge q. Lets now connect
a battery between the two boxes with an EMF ∆V . We can assume that the chemical potential is fixed. If
we add a potential to them the chemical potential will increase. If we consider a few of the energy levels
of the system. If we now add a potential than we simply raise all of the energy levels, so that subsequently
we have a situation where we have simply increased the energy of all the particles by an amount q∆V .
We thus have both an internal and external contribution to the chemical potential. We apply a voltage
difference ∆V to cancel ∆µ, thus
q∆V = ∆µ(initial)
92
we can see that
µ1 ( 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 potential
is nothing more than a potential energy. So the total chemical potential is
µtot = µint + µext
where µext is the potential energy of the particle due to an external source of energy and µint is the energy
per particle without the external energy. The externla contribution could be, electric, magnetic, gravitatinal, etc. The total potential is often referred to as the electric-chemical potential if the external contribution
comes 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). The
hydrogen 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 + + H2 SO4 + 2e− → PbS04 + 2H2 O
anode
the net result is that we use up the lead sulfate and produce water. The electrons come from the external
circuit. The cathode is a sink for sulfate ions, therefore the chemical potential is lower at the cathode (lead
−−
surface) µ(SO−−
4 ) < µ(SO4 ) 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 for the hydrogen atoms. So that the chemical
potential µ(H +) at Pb surface is less than µ(H + ) in electrolyte. Thus there is a step at ∆µ(H + ). The
chemical potential steps drive the ions toward the electrodes and hence the current through the resistor,
or more generally, your external circuit. What happens if we disconnect the external circuit? The current
stops, obviously, how can we stop the chemical reactions? What will happen is that electrons will leave the
anode into the electrolyte until they build up a potential difference that exactly cancels chemical potential
and 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 usually
referred to as the half-cell potentials, where
and
∆V− = −0.4V
∆V+ = 1.6V
∆V = 2.0V
We can see that the battery takes chemical energy to drive electric circuits. The electric chemical potential
is zero to produce zero current flow. The charging process reverses the chemistry and discharge process
and it takes sulfate ions and produces sulfuric acid. Removes SO4 from the electrodes to produce sulfuric
acid.
93
5.5 The Gibbs Factor
We know that τ gives us the Boltzman factor and we will now see that the chemical potential gives us the
Gibbs factor. We assume that we have a resevior in thermal contact with a system. Where the number of
particles in the resevoir is N0 − N and energy u0 − εl . The system is diffusive and in thermal contact. We
want to find the ensemble average, where the system has N particles with energy εl,N . We will start of by
assuming that there is only one state that has energy εl so that we have a non-degenarate system. Thus the
multiplicity is equal to 1. We can say that the probability that this is in some state εl with N particles is
proportional to multiplicity of the resevoir
P ∝ g(N0 − N, u0 − εl ) × 1
i.e
P(N, εl ) = Ag(N0 − N, u0 − εl )
we can now do a first order Taylor expansion
∂ ln g
ln g(N0 − N, u0 − εl ) = ln g(N0 , u0 ) −
∂N
∂ ln g
N−
∂u
u0
thus we can see that
ln g(N0 − N, u0 − εl ) = ln g(N0 , u0 ) +
εl
N0
µN εl
−
τ
τ
if 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 the
chemical potential. We can now make a couple of immediate conclusions. You first need the ratios of the
probability
P(N1 , ε1 ) e(N1 µ−ε1 )/τ
=
P(N2 , ε2 ) e(N2 µ−ε2 )/τ
we sometimes get degeneracies, where we can use the same trick as for the Boltzman factor. If we suppose
that we have energy ρ1 and ε1 and ρ2 and ε2 . Thus the ratios of the probabilities of the system having
energies ε1 and ε2 is given by
W (N1 , ε1 ) ρ1 e(N1 µ−ε1 )/τ
=
W (N2 , ε2 ) ρ2 e(N2 µ−ε2 )/τ
where we now weight the factors.
What is B? We need to sum
N0
∑
∑ P(N, εl ) = B
N=0 εl
N0
∑ ∑ e(Nµ−εl )/τ = 1
N=0 εl
thus
B=
1
ζ
where
ζ=
94
N0
∑ ∑ e(Nµ−εl )/τ
N=0 εl
this we call the Grand Partition Function or the Grand Sum. Finally we have the results for the Gibbs
factor
e(Nµ−εl )/τ
P(N, εl ) =
ζ
this is known as the Gibbs factor it resembles the the Boltzman factor. We want to find the average number
of particles and the average energy of the system.
Mean Number of particles hNi
We know that the average nuber of particles is given as
hNi =
∑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 some
algabreac manipulation. Consider
!
∂ζ N
ζhNi
1
N (Nµ−εl )/τ =
= ζ=
∑
∑
∂µ
τ
τ N l
τ
what we see is that the average value of N is given by
hNi =
∂ ln ζ
τ ∂ζ
=τ
ζ ∂µ
∂µ
we will also calculate the average energy. The mea energy hεi
hεi =
∑N ∑l εl ∑l Neβ(Nµ−εl )
ζ
lets consider the following derivative
∂ζ
= ∑(Nµ − εl )eβ(Nµ−εl ) = [hNiµ − hεi]ζ
∂β ∑
N l
thus
hNiµ − hεi =
1 ∂ζ ∂ ln ζ
=
ζ ∂β
∂β
and we find the average energy to be given as
hεi = hNiµ −
∂ ln ζ
∂ ln ζ ∂ ln ζ
= µτ
−
∂β
∂µ
∂β
and finally
hεi = µτ
∂ ln ζ ∂ ln ζ
−
∂µ
∂β
a result from all this is for the fluctuations
h(∆N)2 i = τ
this is left as a homework assigment.
95
∂hNi
∂µ
5.6 Summary
This are some very important relationships
∂σ
∂σ
∂σ
dµ +
dV +
dN
dσ(u,V, N) =
∂µ V,N
∂V µ,N
∂N V,µ
µ
du P
+ dV − dN
=
τ
τ
τ
this gives us the following identity
dµ = 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
τ
p
µ
σ(u,V, N)
∂σ
1
=
τ
∂µ
V,N
p
∂σ
τ = ∂V µ,N
µ
∂σ
− τ = ∂N
u(σ,V, N)
∂µ
τ = ∂σ
V,N
∂µ
−p = ∂V
σ,N
∂µ
µ = ∂N
σ,V
µ,V
We then talked about the total chemical potenial
µtot = µint + µext
we then talked about the Gibbs factor
P(N, εl ) =
where ζ is the grand sum
e(Nµ−εl )/τ
ζ
ζ = ∑ ∑ e(Nµ−εl )/τ
N
l
and we found the average number of particles
hNi =
∂ ln ζ
τ ∂ζ
=τ
ζ ∂µ
∂µ
and the average energy is given as
hεi = µτ
∂ ln ζ ∂ ln ζ
−
∂µ
∂β
and the fluctuations are given by
h(∆N)2 i = τ
96
∂hNi
∂µ
F(τ,V, N)
−p = ∂F
∂V
τ,N
∂F
µ = ∂N
τ,V
5.7 Problems and Solutions
Problem # 1 States of positive and negative ionization
Consider a lattice of fixed hydrogen atoms; suppose that each atom can exist in four states:
state
Numbe o f electrons Energy
ground
1
− 21 ∆
Positive ion
0
− 21 δ
1
Negative ion
2
2δ
1
Excited
1
2∆
Find the condition that the average number of electrons per atom be unity. The condition will involve
δ, λ, and τ.
The grand partion function is given as
Z = ∑ e(Nµ−εn )/τ = e(µ+ 2 ∆)/τ + e 2 δ/τ + e(2µ− 2 δ)/τ + e(µ− 2 ∆)/τ
1
1
if we let
1
1
λ = eµ/τ
which is the absolute activity, allows us to simplify the partition function to
Z = e 2 ∆/τ λ + e 2 δ/τ + e− 2 δ/τ λ2 + e− 2 ∆/τ λ
1
1
1
1
Z = λ[e 2 ∆/τ + e− 2 ∆/τ ] + e 2 δ/τ + e− 2 δ/τ λ2
1
1
1
1
we know that the average number of particles is given by
i
i
1
1
1
∂ h h 1
∂
hNi = λ ln Z = λ ln λ e 2 ∆/τ + e− 2 ∆/τ + e 2 δ/τ + e− 2 δ/τ λ2
∂λ
∂λ
we know that
i
1
1
1
e 2 ∆/τ + e− 2 ∆/τ + 2λe− 2 δ/τ
∂ h 1 ∆/τ
ln λ[e 2 + e− 2 ∆/τ ] + e 2 δ/τ + e− 2 δ/τ λ2 =
1
1
1
1
∂λ
λ[e 2 ∆/τ + e− 2 ∆/τ ] + e 2 δ/τ + e− 2 δ/τ λ
1
1
1
thus we find that the average number of particles is
"
#
1
1
1
e 2 ∆/τ + e− 2 ∆/τ + 2λe− 2 δ/τ
hNi = λ
1
1
1
1
λ[e 2 ∆/τ + e− 2 ∆/τ ] + e 2 δ/τ + e− 2 δ/τ λ2
setting this expression equal to 1 yields
"
#
1
1
1
e 2 ∆/τ + e− 2 ∆/τ + 2λe− 2 δ/τ
λ
=1
1
1
1
1
λ[e 2 ∆/τ + e− 2 ∆/τ ] + e 2 δ/τ + e− 2 δ/τ λ2
thus
λ[e 2 ∆/τ + e− 2 ∆/τ + 2λe− 2 δ/τ ] = λ[e 2 ∆/τ + e− 2 ∆/τ ] + e 2 δ/τ + e− 2 δ/τ λ2
1
1
1
1
97
1
1
1
2λ2 e− 2 δ/τ = e 2 δ/τ + λ2 e− 2 δ/τ
1
1
we find
1
λ2 e− 2 δ/τ = e 2 δ/τ
1
thus
1
λ = eδ/2τ
Problem # 2 Carbon monoxide poisoning
In carbon monoxide poisoning the CO replaces the O2 adsorbed on hemoglobin (Hb) molecules in the
blood. To show the effect, consider a model for which each adsorption site on a heme may be vacant or
may be occupied either with energy εA by one molecule O2 or with energy εB by one molecule CO. Let N
fixed heme sites be in equilibrium with O2 and CO in the gas phases at concentrations such that the
activities are λ(O2 ) = 1 × 10−5 and λ(CO) = 1 × 10−7 , all at body temperature 370 C. Neglect any spin
multiplicity factors.
a) First consider the system in the absence of CO. Evaluate εA such that 90 percent of the Hb sites are
occupied 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 ) =
thus we can see that
which simplifies to
λe−εA /τ
= 0.9
1 + λe−εA /τ
λ = 0.9(eεA/τ + λ)
λ
ln
− λ τ = εA = −13.71τ = −13.71 × kB T
0.9
since we know that T = 310 K thus we find that the energy is
εA = −.364 eV/O2
b) Now admit the CO under the specified conditions. Find εB such that only 10 percent of the Hb sites
are occupied by O2 .
98
In this case we know that the grand partition function can be expressed as
Z = 1 + λO2 e−εA /τ + λCO e−εB /τ
where the first term is for an unoocupied state, the second term is when it is occupied by O2 and the third
term is for when it is occupied by CO. thus we can see that the probability is given by
P(N, εA ) =
thus we find that
λO2 e−εA /τ
= 0.1
1 + λO2 e−εA /τ + λCO e−εB /τ
λO2 e−εA /τ = 0.1[1 + λO2 e−εA /τ + λCO e−εB /τ ]
we can see that the energy is given by
εB = −τ ln
""
#
#
1
λO2 e−εA /τ
− 1 − λO2 e−εA /τ
0.1
λCO
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
τ ∂ζ
hNi =
ζ ∂µ τ,V
from
hNi =
∂ ln ζ
τ ∂ζ
=τ
ζ ∂µ
∂µ
a) Show that
hN 2 i =
τ2 ∂2 ζ
ζ ∂µ2
the mean-square deviation h(∆N 2 )i of N from hNi is defined by
h(∆N)2 i = h(N − hNi)2 i
= hN 2 i − 2hNihNi + hNi2
= hN 2 i − hNi2
"
2 #
2
1 ∂ ζ 1 ∂ζ
h(∆N)2 i = τ2
−
ζ ∂µ2 ζ2 ∂µ
since we know that
hN 2 i = h(∆N)2 i + hNi2
99
(5.1)
we can see that
h(∆N)2 i = τ2
"
1 ∂2 ζ 1
−
ζ ∂µ2 ζ2
and also we can see that
hNi2 =
∂ζ
∂µ
2 #
τ2 ∂2 ζ
ζ2 ∂µ2
putting this into equation 1 we find
hN 2 i = τ2
"
1 ∂2 ζ 1
−
ζ ∂µ2 ζ2
∂ζ
∂µ
2 #
+
τ2 ∂2 ζ
ζ2 ∂µ2
doing the math yields the solution we were looking for
hN 2 i =
τ2 ∂2 ζ
ζ ∂µ2
b) Show that this may be written as
∂hNi
∂µ
in Chapter 6 we apply this result to the ideal gas to find that
h(∆N)2 i = τ
1
h(∆N)2i
=
hNi2
hNi
is the mean square fractional fluctuation in the population of an ideal gas in diffusive contact wit a
resevoir. If hNi is of the order of 1020 atoms, then the fractional fluctuation is exceedingly small. In
such a system the number of particles is well defined even though it cannot be rigorously constant
because diffusive contact is allowed with the resevoir. When hNi is low, this relation can be used
in the experimental determination of the molecular weight of large molecules such as DNA of MW
108 − 1010 .
Since we know that
hNi =
where we can see that
∂
∂hNi
=
∂µ
∂µ
using the product rule we find
τ ∂ζ
ζ ∂µ
τ ∂ζ
ζ ∂µ
∂hNi τ ∂2 ζ τ ∂2 ζ
=
−
∂µ
ζ ∂µ2 ζ2 ∂µ2
thus we find that
h(∆N) i = τ
2
thus we have just shown that
2
τ ∂2 ζ τ ∂2 ζ
−
ζ ∂µ2 ζ2 ∂µ2
2
τ ∂2 ζ
∂hNi
2 τ∂ ζ
=τ
−
h(∆N) i = τ
∂µ
ζ ∂µ2 ζ2 ∂µ2
2
100
Problem # 4 Multiple binding of O2
A hemoglobin molecule can bind four O2 molecules. Assume the ε is the energy of each bound O2 ,
relative to O2 at rest at infinite distance. Let λ denote the absolute activity eµ/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 the
result qualitatively as a function of λ.
We know that the grand partition function for this system is given by
Z=
∑
e(Nµ−εs(N) )/τ = 1 + 4e(µ−ε)/τ + 6e2(µ−ε)/τ + 4e3(µ−ε)/τ + e4(µ−ε)/τ
ASN
This is because we know that if there are no molecules attached then we get the first term, the second term
is assuming that there is one oxygen atom attached we find the multiplicity to be 4, if there are two atoms
attached then the multiplicity is 6, if there are 3 atoms attached then the multiplicity is 4, and if there are
four atoms attached then the multiplicity is 1. Since we know that the absolute activity is defined as
λ = eµ/τ
allows us to write the grand partition function as
Z = 1 + 4λe−ε/τ + 6λ2 e−ε/τ + 4λ3 e−ε/τ + λ4 e−ε/τ
we know that the probability that one atome is attached is given by
P(N = 1, ε) =
4λ
4λe−ε/τ
= ε/τ
−ε/τ
2
−ε/τ
3
−ε/τ
4
−ε/τ
1 + 4λe
+ 6λ e
+ 4λ e
+λ e
e + 4λ + 6λ2 + 4λ3 + λ4
The plot is given
0.04
P(λ)
0.03
0.02
0.01
0.00
0
200
400
600
λ
800
1000
1200
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 particles
that cannot be labeled, they are strictly identical. We can write down a Shrodinger equation that is time
independent
~2
~2 2
− ∇21 ψT −
∇ ψT + (V1 +V2 )ψT = ET ψT
2m
2m 2
we are assuming that this particles are i some electric field that give them the potential energy. We can
solve by writing down the wave functios by bte product of the two wave functionc
ψT = ψα (1)ψβ(2)
where α and β are te quantum numbers of each particle. If we plug this in we get the following result
−
~2
~2
ψβ (2)∇21 ψα (1) −
ψα (2)∇22 ψβ (2) + (V1 +V2 )ψα (1)ψβ (2) = ET ψα (1)ψβ(2)
2m
2m
we can see that this two equations seperate. This can be written as
−
~2 ∇21 ψα (1)
+V1 = E1
2m ψα (1)
−
~2 ∇22 ψβ (2)
+V2 = E2
2m ψβ (2)
103
ET − E1 + E2
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 label them, it only allows us to distinguish their
states. We cannot label identical particles, so an equaly good wave function is of the form ψα (2)ψβ(1) so
the general solution is of the form
1
ψT = √ [ψα (1)ψβ(2) ± ψβ (1)ψα (2)]
2
this brings us to the question of the symmetry of the wave function, symmetric and anti-symmetric. The
consequencies of this is very profound. What we learn emperical is that ther are two general classes of
behaviours. 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 integer
spins, i.e photons, deuterons, and helium 4 nucleus. So the first class, the fermions have anit-symmetric
wave functions and the bosons have symmetric wave functions. What difference does this make? If
we interchange the symmetric wave functions we can see that there is no change, on the other hand
interchanging the anti-symmetric wave functions we can see that there is a change ψ− → −ψ− . What
happens if we let α = β. For the symmetric case, there is nothing that is violated, the total wave function
simply becomes
2
ψT = √ ψα (1)ψα (2)
2
if we do this for the anti-symmetric case we see that the wave function vanishes. This is a result of the
Pauli exclusion principle, you cannot have two identical particles in the same quantum state. There is no
restriction for bosons.
6.2 Fermi-Dirac Distribution
We will use the grand canonical ensemble, we must consider a a lot 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 is
to find the average occupancy that has an energy εl . Since these are fermios means that we can only have
1 particle in a given state. We know that N = 0 or N = 1. This system is in thermal equilibriums that can
exchange energy and particles. We want to find the average number of particles in this system.
hN(εl )i
we must write down the grand partition function.
ζ = 1 + e(µ−εl )/τ
these are the only two allowed states
hN(εl )i =
e(µ−εl )/τ
1 + e(µ−εl )/τ
this is know as f (ε)
f (ε) =
1
e(ε−µ)/τ + 1
this is known as the Fermi-Dirac distribution function. We can see that this result involves the potential
energy. 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 at τ = 0? If we take energy
ε < µ
e(ε−µ)/τ = e−∞ = 0
ε > µ
e(ε−µ)/τ = e∞ = ∞
f (ε < µ) = 1
f (ε < µ) = 0
if we were to plot this we would get a step function at µ. Secondly, at any temperature
f (ε = µ) =
1
1
=
e0 + 1 2
thus µ is the energy for which the occupancy is 1/2. What happens with increasing temperature? Lets
suppose that we have a 3-D gas with a Fermi gas. Lets take spin 1/2 with a lot of energy levels and we
also have an even number of Fermions. The lowest energy level will have only two electrons (exclusion
principle). The next two will go into the first excited state etc. We will see that the electrons in the higher
energy state will move to even higher energy states. We will smear out the distribution around the energy
levels τ(kB T ). The Fermi level is εF , and at T = 0 it is exactly the same as the chemical potential µ. This
smearing is symmetric around the chemical potential.
Symmetry of Filled and Vacant states
What is the probability that the state µ = δ is
f (µ − δ) =
1
f (µ + δ) =
e−δ/τ + 1
1
eδ/τ + 1
this is the probability that these two states will be occupied. What is the probability that at µ + δ is vacant?
(eδ/τ + 1) − 1
eδ/τ
1
p(µ + δ) = 1 − f (µ + δ) =
= δ/τ
= −δ/τ
= f (µ − δ)
δ/τ
e +1
e +1 e
+1
if we now imagine what happens to this distribution as we increase the temperature. We can see that there
will be a smearing of the dsitribution that is of the order of τ. The chemical potential will eventually start
to change, as τ is increased than µ 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 put as many bosons as we like in any
quantum state. As before we consider 1 quantum state l that is in thermal equilibrium with all the other
states in the resevoir. We ca now write down the grand partition function
ξ = 1 + e(µ−εl )/τ + 2e2(µ−εl )/τ + ...
both µ and ε get multiplied by the number
ξ=
N0 →∞
∑
en(µ−εl )/τ
n=0
105
we will assume that N0 → ∞. If we now let
x = e(µ−εl )/τ
thus we find
ξ=
∞
1
∑ xn = 1 − x
n=0
this only works if x < 1, otherwise this function blows up. We can see that
µ − εl
<0 µ<ε
τ
what we now want to calculate the average occupancy of the state l.
hn(ε)i =
6 probability of state being occupied
this is given as
1
d
d
∑ nxn x dx ∑ xn x dx 1−x
hn(ε)i =
=
=
=
1
∑ xn
∑ xn
1−x
x
(1−x)2
1
1−x
=
1
x−1
thus we find the Bose-Einstein distribution
hn(ε)i = n(ε) =
1
e(ε−µ)/τ − 1
the Bose-Einstein distribution will always have ε > µ. The Bose-Einstein distribution blows up as e(ε−µ)/τ →
1, this function converges the the Fermi-Dirac distribution when ε − µ ≫ τ. The Bose-Einstein condensation refers to the the fact that at low enough temperatures all of the bosons will condense into the ground
state.
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
mτ 3/2
1
≈ 3
nQ =
2
2π~
λD
thus for 2 particles seperated by more than the Debroigle wavelength, we find that the concentration of
the particles n ≈ nQ . If we have two particles that are very close with respect to the Debroigle wavelength
we must consider the quantum effects and must utilize quantum statistics. If we have two particles that
are seperated by much more than the Debroile wavelength than we 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 =
∑ e−εn/τ = nQV
n=0
where n2 = n2x + n2y + n2z and V 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 is
one in which f (ε) or n(ε) is ≪ 1. This is a low density level, at this level, both distributions give the same
results. So that
e(ε−µ)/τ ≫ 1
106
thus we find
n(ε) = eµ/τ · e−ε/τ
this is the distribution function for a classical gas. We want to now figure something out about the chemical
potential. Lets confine our attention to a monotomic gas, i.e He. What is µ? We can find µ from the
following
N = hNi = ∑ f (εn )
n
we simply add the occupancies of all the states. We find
N = eµ/τ ∑ e−εn /τ = eµ/τ Z1 = nQVeµ/τ
n
this is rather neat, we can devide by the volume on both sides to get
n
N
= eµ/τ
= nQ eµ/τ ⇒
V
nQ
taking logs on both sides
n
µ = τ ln
nQ
this links the classical potential µ to the densities. If we we look at our previous expression
n
µ/τ −ε/τ
n(ε) = f (ε) = e e
=
e−ε/τ
nQ
thus f (ε) ≪ 1 provided n ≪ nQ . We can see from the expression of the chemical potential that for a
classical gas, the chemicla potential is negative. This is yet another signature between the classical gas
and the quantum gas. If we lower n/nQ we can see that the chemical potential changes signs at n = nQ ,
where we go from classical to quantum regimes. The definition of the chemical potential always involves
nQ which always involve ~. Thus even at the classical limit, the particles “remember” that they are made
up 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 that τ is fixed. If we consider two layers of gas that are in diffusive contact,
which allows particles and energy to be transferrable through the layer. The layer is filled with monotomic
particles in thermal equilibrium. If we also assume that we have zero potential energy at sea level. We can
see that the particles above sea level will have
µext = mgd
thus the total chemical potential is
µtot
n
= τ ln
nQ
107
+ mgd
now, in thermal equilibrium, the total chemical potential must be constant. We can see that as we go up
in altitude, µext increases and µint will decrease. We can make the following statement: if we consider the
upper level
n(0)
n(d)
+ mgd = τ ln
τ ln
nQ
nQ
combining the two log terms we find
n(d)
τ ln
n(0)
= −mgd
that is to say that
n(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)
where d(0) = τ/mg, if we put some numbers, i,.e N2 we find
d(0) =
290 K
≈ 8.7
28 × 1.7 × 10−27 kg × 9.8 m/s2
6.5 Ideal Classical Gas:Free Energy, Pressure, Internal Energy, Entropy, and Heat Capacity
The word ideal means that there are no interaction between the particles, and classical mean that we are
in the low energy limit. The free energy is the most important quantity one will derive. Recall that the
chemical potential µ is
Mτ 3/2
µ = τ ln(n/nQ ) nQ =
2π~2
the first thing we need to find is the free energy
6.5.1 Free Energy
µ=
where
∂F
∂N
τ,µ
3
M
3
µ = τ ln N − lnV − ln τ − ln
2
2 2π~2
this is for later convenience, so we can then write
F(N, τ,V ) =
Z N
0
dNµ(N, τ,V ) = τ
Z N
0
3
3
2π~2
dN ln N − lnV − ln τ + ln
2
2
M
lets recall the following result
Z
ln xdx = x ln x − x
108
we get the following
3N
3N
2π~2
F(N, τ,V ) = τ N ln N − N − N lnV −
ln τ +
ln
2
2
M
the first two terms give you
N ln n = N ln N − N − N lnV
and the last two terms give you
3N
2π~2
3N
ln τ +
ln
−N ln nQ =
2
2
M
thus
F(N, τ,V ) = Nτ ln n − Nτ ln nQ = Nτ[ln n − ln nQ ]
thus
n
−1
F(N, τ,V ) = Nτ ln
nQ
this is the free energy.
6.5.2 Pressure
The pressure is defined as
∂F
p=−
∂V
τ,N
using Equation 6.1 we find
Nτ
p=− −
V
thus we get the ideal gas law
pV = Nτ = NkB T = RT
where R is the gas constant for 1 mole. We can immediatly figure out that
R = NA kB = 6.02 × 1023 atoms/mole × 1.38 × 10−23 J/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
thus the internal energy is
u = F + τσ
∂F
σ=−
∂τ
∂F
u = F −τ
∂τ
109
V,N
(6.1)
there is a trick, lets consider
!
1
F
∂F
∂F
2
2 ∂
= −τ − 2 F +
τ = F −τ
−τ
∂τ τ
τ
∂τ V,N
∂τ
thus
u = −τ
2
∂
∂τ
F
τ V,N
but from Equation 6.1 we can see that there is only one term in τ, thus
3N 1 3
2
u = −τ −
= Nτ
2 τ 2
which is the same result we have derived before from the equipartition theory.
3
u = Nτ
2
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
∂F
σ=−
∂τ V,N
once again looking at Equation 6.1, and differentiating with respect to τ, hence
3N 1
3
3
2π~2
σ = − N ln N − N − N lnV − Nτ + N ln
−τ
2
2
M
2 τ
where the last term came from
∂
∂τ
3N
ln τ
2
thus we can see that the entropy is given by
5
3
3
2π~2
+ N
σ = − N ln n − N ln(τ) + N ln
2
2
M
2
we also know that
3
3
2π~2
− ln(nQ )N = − Nτ + N ln
2
2
M
putting all these together we have
nQ
5
σ = N ln
+
n
2
this result is called the Sackus-Tetrode Equation. This is the central result of the ideal gas, which will
allow you to derive all the other quantities. Note: Gibbs paradox (∼ 1875) was the first indication of
problems with classical physics (indistinguishibility)
110
6.5.5 Heat Capacity
We have
du = τdσ − pdV
where the first term is the heat added and the second term is the work done on the gas. There are two kinds
of heat capacity
• Heat capacity at constant volume. We have a box with a bunsen burner, the heat increases but the
volume stays the same
• Heat capacity at constant pressure. We have a piston with pressure being applied, along with a
bunsen burner that adds heat to the system.
In the first case there is no mechanical work, and in the second case there is mechanical work that is
applied in pushing back the piston against constant pressure. We can see that
CP > CV
lets start with CV . If we want to increase the temperature dτ then the internal energy must increase by du.
The internal energy depends only on the temperature. We can see
δu
δσ
=τ
δτ V
δτ V
thus
CV =
∂u
∂τ
V
∂σ
=τ
∂τ
V
we rember that the entropy is
3
5
=N
ln τ + terms independent of τ
σ = N ln nQ − ln n +
2
2
where we can see that
∂σ
∂τ
3 1
= N
2 τ
thus the heat capacity is
3
3
CV = N = NkB
2
2
now for the heat capacity at constant pressure CP , we begin with
τdσ = du + pdV
thus
∂σ
CP = τ
∂τ
we can now write this as follows
CP =
∂u
∂τ
P
∂V
+p
∂τ
P
111
the first term is once again CV , that is because the internal energy depends only on the temperature. The
second term is the work done in expanding the gas, which we must figure out, thus
∂V
C p = CV + p
∂τ P
thus we can see that
pV = Nτ
thus
thus
∂V
p
∂τ
V=
=p
P
Nτ
p
N
=N
p
5
5
CP = CV + N = N = NkB
2
2
and we can see that
CP > CV
this is only for a monotomic gas. The final thing is to define the ratio of CP to Cv where
γ=
CP 5
=
CV
3
6.6 Expansion of Gases
There are two kinds of expansion in gases, one in which we keep the temperature fixed and the other where
we 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 some
resevoir that we keep at constant temperature T0 , we then have some cylinder with a piston that applies a
constant pressure on the system. What we know is that pV = Nτ and since we know that the temperature
is constant we know that piVi = p f V f , this is known as Boyle’s Law. What we know is
nQ
5
σ = N ln
+
= N lnV + terms independent of V
n
2
in this context the entropy only depends on the first term. The change in the entropy is given as
Vf
∆σgas = N lnV f − N lnVi = N ln
Vi
if the final volume is bigger then the initial volume then we can see that the entropy increases. What is the
work done on the gas?
Z Vf
Z Vf
Z Vf
F
pdV
dV = −
Fdx = −
∆W = −
Vi
Vi A
Vi
going back to the ideal gas law we find
Z Vf
Vf
Nτ
∆W = −
dV = −Nτ ln
Vi
Vi V
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 the
internal energy
3
u = Nτ
2
thus
∆u = 0
from this we can talk about the heat flow from or into the gas, if the change in the internal energy is zero
we 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 do work to make this happen. This is
part of the thermal cycle of a heat engine. This is the heat flow during expansion. The final thing we can
ask is what is the change i the entropy of the resevoir, this is given by
∆Qgas ∆W
∆Qgas
=−
=
= −∆σgas
τ
τ
τ
change in entropy of resevoir =
and we can see that
∆σres + ∆σgas = 0
which implies
∆σ = 0
this implies that this is an entirely reversible process. We will 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 thermodynamics
the word adiabatic means that it is thermally isolated. Conceptually, we have a cylinder and a piston with
gas atoms inside, but there is no heat exchange with the system and the resevoir. When we change the
volume Vi → V f there is no heat exchange (∆Q = 0 and ∆σgas = 0) with the environment and thus this is
an adiabatic process. We want to find a relationship between the pressure and the volume. Lets figure out
what this is, we will end up using CP ,CV and γ = CP /CV . We will begin using the thermodynamic identity
du = τdσ − pdV
we know that dσ = 0 and we also know
CV =
∂u
∂τ
V
113
du = CV dτ
and we can see that
CV dτ + pdV = 0
we need to eliminate dτ, we can use the ideal gas equation
d(pV ) = d(Nτ)
pdV +V d p = Ndτ
we can see
CV (pdV +V d p) + N pdV = 0
we can rewrite this as follows
pdV (CV + N) +CV V d p = 0
we can see that this simplifies to
CP
pdV +V d p
CV
γpdV +V d p
dV d p
γ
+
V
p
γ lnV + ln p
which gives
= 0
= 0
= 0
= C
pV γ = Constant
this is the result used for adiabatic expansions. Also, lets consider
pV = Nτ
we can see that
N
τ
V
τV γ−1 = Constant
or
Vγ =
which yields
p=
N γ τγ
pγ
pγ−1 τγ = Constant
the last thing we should work out is the work done on the gas. We use
∆W = u f − ui
this is simply
3
N(τ f − τi )
2
τf
3
−1
Nτi
=
2
τi
!
1−γ
Vf
3
=
−1
Nτi
2
Vi
∆W =
114
which gives
3
∆W = Nτi
2
"
Vf
Vi
−2/3
−1
#
when the gas expands there is a increase in σ due to the increase in the volume. If the temperature drops,
this gives a decrease in σ which exactly cancels the increase in σ 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 expand
the box very slowly so that the energy levels are all lower. The result of doing time dependent pertubation
tells us that the atoms do not change energy levels, and thus the distribution of the particles over the
quantum 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 all
the atoms are confined to the left hand side, we than remove the partition very suddenly. We thus go into a
new state in whihc the particles are distributed uniformly over the volume of the box. We can see that this
is an irreversible process, i.e placing the partition back in will not give us the same distribution. First of
all, 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 the
change in the entropy is not zero, it is
Vf
∆σ = N ln
Vi
this is analogous to what happens in QM, we can compare this to the sudden approximation in QM, i.e
beta 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 energy kT to excite
the vibration. We will imagine a set of distinct energy levels given by
1
ε = n~ω n = 1, 3, 5, ...
2
in the classical limit we know that u = kB T thus the heat capacity is CV = kB , this of course is in the limit
kB T ≫ ~ω. What about rotation? There are two degrees of rotation for polyatomic gases, thus the energy
of a rotator is given as
1
urot = Iω2
2
where I is the moment of inertia, there is nonly a kinetic energy term. We can also write this as
urot =
1 (Iω)2
2 I
115
in QM we have the following
h(Iω)2i = r(r + 1)~2
where r is an integer and Iω is the angular momentum. Thus the rotational energy is given by
urot =
r(r + 1)~2
2I
r = 0, 1, 2, 3, ...
once again since this is in the classical limit we will have 12 kB T for each degree of rotation, thus
urot (2 modes) = kB T
and the heat capacity will be
CV = kB
the rotation mode about the axis through the atoms is never excited, it would take an infinite amount of
energy since the moment of inertia is approximately 0. If we were to go to NH3 we would have 3 modes
of rotation. Given this result we can now ask how big is the rotational energy.
Rotational energy of N2
Lets consider the seperation between the center of rotation and one atome be 10−10 meters. What is the
moment of inertia, we know that MN2 ≈ 2 × 10−26 kg, we find the moment of inertia to by
I = 2 × (10−10 )2 × 2 × 10−26 kg ≈ 4 × 10−26 m2 kg
thus the energy of the first excited state
u=
2 × (10−34 )2
≈ 2 × 10−22 J ≈ 10 K
2 × 4 × 10−26
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 of
Freedom
This is used when the problem implies that there are internal degrees of freedom. The grand sum for
bosons and fermions become the same in the classical limit. For he case o the fermions when you have
two energy levels you can have at most one particles in the ground state. For fermions
ζ = 1 + e(µ−εn )/τ
for bosons we have
ζ = 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 grand
partition functions are the same. Since f (εn ) ≫ 1 we can neglect the higher order terms. Hence in the
classical 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 of
freedom. For the polyatomic case εn → εn + εint , where this corresponds to the vibrational and rotational
energy, thus we now have
ζ1 + ∑ e(µ−εn −εint )/τ
int
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 factor Zint , which yields
ζ = 1 + Zint e(µ−εn )/τ
we can use this to find a new expression for the chemical potential. The probability that the state n is
occupied for any state of its internal motion is given by the following
f (εn ) =
Zint e(µ−εn )/τ
1 + Zint e(µ−εn )/τ
but now by hypothesis we have agreed that the grand partition function is very small, thus we find
f (εn ) = Zint e(µ−εn )/τ
this is the result for the monotomic gas multiplied by the factor representing the internal degrees of freedom. We can now relate this to the average occupancies
N = hNi = ∑ f (εn ) = Zint eµ/τ ∑ e−εn /τ = Zint eµ/τ Z1
which gives
N = Z1 Zint eµ/τ
but we remember that
Z1 = nQV
we can now write
N = nQV Zint eµ/τ
rearranging gives us the following
n
= Zint eµ/τ
nQ
taking the log of both sides give
n
µ = τ ln
− ln Zint
nQ
we can see that we now have an additional term that accounts for the internal degrees of freedom, without
this 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 to a high enough temperature to ignite the diesel
fuel that is then injected. If we start we air at T = 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
γ−1
γ−1
TiVi
= Tf V f
T f = Ti
thus
Vi
Vf
γ−1
we know that γ = 7/5 for air, which is nitrogen and oxygen, this is because
3
CV = NkB + NkB
2
where the second term is added for the rotational mode, and we know that
7
CP = CV + NkB = NkB
2
and
γ=
7
5
if we put the number in we get
T f = 300(16)2/5 = 909 K
6.7.1.2 Example 2
We need to show that
p=
3U
2V
this is done in three steps.
a) Find that the average pressure in a system in thermal contact with a heat resevior. Recall
∂εs
ps = −
∂V N
thus the average pressure is equal to
hpi = ∑ ps P(εs )
s
where p is the pressure and P is the probability that state s is occupied. We find
−εs /τ
e
∂εs
hpi = − ∑
∂V N Z
s
this is step 1.
118
b) Show that for an ideal gas
∂εs
∂V
N
=−
2 εs
3V
there are various ways of getting this. If we have a cube of side L we find
1
~2 π 2 2
(nx + n2y + n2z ) ∝ 2/3
εs =
2
2m L
V
for all nx , ny , and nz . If we take logs of both sides we find
2
ln εs = − lnV
3
if we differentiate both sides we find
dεs
2 dV
=−
εs
3 V
which gives
∂εs
∂V
N
=−
2 εs
3V
c) the result of the pressure
2
2 εs −εs /τ
1
e
=
p=− ∑ −
Z s
3V
3V
∑s εs e−εs /τ
Z
!
=
2U
3V
if we go for a monotomic gas, the energy U = 23 NkB T we can see that
pV = NkB T
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/τ e−ε/τ = λe−ε/τ
where λ is the absolute acitivity. we also found
µ = τ ln(n/nQ )
also
F = Nτ[ln(n/nQ ) − 1]
and
pV = Nτ
119
also the internal energy
3
U = Nτ
2
and the entropy is
the heat capacity is
5
σ = N ln(n/nQ ) +
2
3
CV = N
2
and
5
C p = CV + N = N
2
we also learned about isothermal expansion
piVi = p f V f
we also saw that
∆σres = −N ln(V f /Vi ) ∆σgas = N ln(V f /Vi )
which gives
∆σ = 0
we also talked about the adiabatic expansion
γ
γ
piVi = p f V f
γ=
CP
CV
where this process is a reversible process. We also talked about the free expansion, this is where we rupture
a membrane in some box. We found that there is no change in the internal energy and no change in the
temperature
∆U = 0 ∆τ = 0 ∆σ = N ln(V f /Vi )
this process is irreversible. We also talked about diatomic gases. We found
3
CV = N + N + N
2
where these are for rotational, vibrational and translational energies. We also find
µ = τ [ln(n/nQ ) − ln Zint ]
we can also remember that the 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. The
values 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 hNi, when the system composed of this
orbital is in thermal and diffusive contact with a reservoir at temperature τ and chemical potential µ.
120
We know that the average occupancy is defined as
hNi =
∑N ∑s Ne(Nµ−εs )/τ
ζ
where ζ is the grand partition function
ζ = ∑ ∑ e(Nµ−εs )/τ
N
(6.2)
s
the average occupancy can also be acquired witha bit of algebreac manipulation, Consider
!
ζhNi
1
∂ζ N
N (Nµ−εl )/τ =
= ζ=
∑
∑
∂µ
τ
τ N l
τ
what we see is that the average value of N is given by
hNi =
∂ ln ζ
τ ∂ζ
=τ
ζ ∂µ
∂µ
thus we simply need to know what the grand sum is. This is given by Equation 1, we find (if we let
λ = eµ/τ )
ζ = 1 + λe−ε/τ + λ2 e−2ε/τ
where the derivative is given as
∂ ln ζ 1 ∂ζ 1 1 −ε/τ 2 2 −2ε/τ
=
=
λe
+ λ e
∂µ
ζ ∂µ ζ τ
τ
thus we can see that the average occupnacy is given as
hNi =
λe−ε/τ + 2λ2 e−2ε/τ
1 + λe−ε/τ + λ2 e−2ε/τ
b) Return now to the usual quantum mechanics, and derive an expression for the ensemble average occupancy of an energy level which is doubly degenerate; that is, two orbitals have the identical energy
ε. If both both orbitals are occupied the total energy is 2ε.
We can write the grand partition function for this system, if we know that s = 0, εs = 0 and s = 1, εs = ε
and this is doubly degenerate, and for the last orbital s = 2, εs = 2ε allows us to write the grand sum as
ζ = 1 + 2λe−ε/τ + λ2 e−2ε/τ
and the ensemble average is given by
"
λe−ε/τ + λ2 e−2ε/τ
hNi = 2
1 + 2λe−ε/τ + λ2 e−2ε/τ
Problem # 2 Energy of Gas of Extreme Relativistic Particles
121
#
Extreme relativistic particles have momentum p such that pc ≫ Mc2 , where M is the rest mass of the
particle. The de Broglie relation λ = h/p for the quantum wavelength continues to apply. Show that the
mean energy per particle of an extreme relativistic ideal gas is 3τ if ε ≃ pc, in contrast to 23 τ for the
nonrelativistic problem.
We know that
h
p
λ=
but we know that
ε = pc
ε=
hc
λ
2L
n
L= λ λ=
2
n
thus
hcn
2L
ε(n) =
We know that the mean energy is given as
U = τ2
∂ ln Z
∂τ
thus we must find the partition function, which is given as
Z = ∑e
−εn /τ
n
π
=
2
Z ∞
0
n2 e−hcn/2Lτ dn
if we let
hcn
2Lτ
x=
dn =
dn
2Lτ
hc
2
2
n =x
2Lτ
hc
2
thus we find that the partitio function is given by
Z=
where
Z ∞
0
2Lτ
hc
3
π
2
Z ∞
0
x2 e−x dx
∞
x2 e−x dx = −x2 e−x − 2xe−x − 2e−x 0 = 2
thus we find that the partition function is
2
∂Z
2 2Lτ
= 3πτ
∂τ
hc
3
U=
τ2 ∂Z 3πτ4
=
= 3τ
Z ∂τ
πτ3
2Lτ
Z=π
hc
thus we find
Problem # 3 Entropy of Mixing
122
Suppose that a system of N atoms of type A is placed in diffusive contact with a system of N atoms of
type B at the same temperature and volume. Show that after diffusive equilibrium is reached the total
entropy is increased by 2N log 2. The entropy increase 2N log 2 is known as the entropy of mixing. If the
atoms are identical (A≡ B), show that there is no increase in entropy when diffusive contact is
established. The difference in the results has been called the Gibbs paradox.
We know that the entropy of an ideal gas is given by
nQ
5
σ = N ln
+
n
2
and we also know that the total entropy before the mixing is just the sum of the individual entropies
which is given by
σT = σA + σB
this is simply
5
5
nQ
nQ
σT = NA ln
+
+ NB ln
+
nA
2
nA
2
we also know
N = NA = NB
nA =
NA
V
nB =
NB
V
n = nA = nB
thus we find that the total entropy is given by
5
nQ
+
σT = 2N ln
n
2
the entropy after mixing can be found by assuming
NT = 2N
n=
N
V
n f = 2n
thus the entropy is now given as
nQ 5
σmix = 2N ln
V +
2N
2
and we find that the change in the entropy is given as
nQ
nQ
5
5
∆σ = σT − σmix = 2N ln
+
− 2N ln
+
n
2
2n
2
thus we can see that this is simply
∆σ = 2N ln 2
Problem # 4 Time for a Large Fluctuation
We quoted Boltzmann to the effect that two gases in a 0.1 liter container will unmix only in a time
10
enormously long compared to 10(10 ) years. We shall investigate a related problem: We let a gas of
atoms of 4 He occupy a container of volume of 0.1 liter at 300 K and a pressure of 1 atm, and we ask how
long 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 this initial condition.
123
We know that the number of accesible states is given by
g = eσ
σ = ln g
where the entropy is defined as
nQ
5
σ = N ln
+
n
2
where the quantum concentration is given as
nQ =
thus we find
where we also know
N=
pV
kB T
Mτ
2π~2
3/2
n=
N
V
3
5
Mτ
N
5
=N
+
ln
− ln
σ = N ln nQ − ln n +
2
2
2π~2
V
2
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
5
MkB T
p
pV 3
+
= 3.69 × 1022
ln
− ln
σ=
kB T 2
2π~2
kB T
2
and the number of accesible states is given by
22
g = e3.69×10
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
1
V2
+ σ1 = N ln
+ σ1
σ2 = N ln
V1
2
where these becomes
1
pV
+ σ1 = 3.52 × 1022
ln
σ2 =
kB T
2
thus we find that the number of accesible states is now given by
22
g2 = e3.52×10
c) For the system in the 0.1 liter container, estimate the value of the ratio
number of states for which all atoms are in one-half of the volume
number of states for which the atoms are anywhere in the volume
We know that this is simply
ratio =
21
g2
= e−1.70×10
g
124
d) If the collision rate of an atom is ≈ 1010 s−1 , what is the total number of collisions of all atoms in the
system in a year? We use this as a crude estimate of the frequency with which the state of the system
changes.
We know that the total number of collisions of all atoms is given by
21
total collisions for all atoms = Nr = 2.415 × 10 × 10
10
π × 107 s
yr
= 7.58 × 1038
collisions
yr
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 in c gives us the probability that all of the atoms
will be in half of the container, and we can see that the time is given by
t = (p f )−1 =
1
pf
where p is the probability that the atoms will be in 1/2 of the container and f is the frequency we found in
part d, so we find that the time is
21
e1.7×10
yrs
t=
7.58 × 1038
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 are N atoms in volume V at temperature τ. Find the (a) chemical potential; (b)
free energy; (c) entropy; (d) pressure; (e) heat capacity at constant pressure.
a) Chemical potential?
We know that the chemical potential is found from the condition that the thermal average of the total
number of atoms equals the number of atoms known to be present. This number must be the sum over all
orbitals of the distribution function f (εs)
N = hNi = ∑ f (εs)
s
The Gibbs sum for this system is given as
ζ = 1 + λe−εn /τ = 1 + λ ∑ e−(εn +εint )/τ = 1 + λZint e−εn /τ
int
where
Zint = ∑ e−εint /τ
int
The probabiliy that the translational orbital n is occupied is given by the ratio of the term in λ to the Gibbs
sum ζ
λZint e−εn /τ
≃ λZint e−εn /τ
f (εn ) =
1 + λZint e−εn /τ
125
if we assume that f (εn ) ≪ 1. Thus we can see that
N = λZint ∑ e−εn /τ
(6.3)
n
where
Zint = ∑ e−εint /τ = 1 + e−∆ε/τ
ε1 = 0 ε2 = ∆ε
int
and
∑ e−εn/τ = nQV
n
we can see from Equation 1 that the chemical potential is given as
n
n
= τ ln
− ln Zint
µ = τ ln
nQ Zint
nQ
which simply becomes
i
h
−∆ε/τ
µ = τ ln(n/nQ ) − ln(1 + e
b) The free energy is defined as
Fint = −Nτ ln Zint = −Nτ ln(1 + e−∆ε/τ )
and the total free energy is given by
Ftot
n
−∆ε/τ
− ln(1 + e
)−1
= Nτ ln
nQ
c) The Entropy is defined as
i
∂ h
∂
∂F
−∆ε/τ
−∆ε/τ
−∆ε/τ
=N
τ ln(1 + e
) = N ln(1 + e
) + τ [ln(1 + e
]
σint =
∂τ V,N
∂τ
∂τ
where
∂
∆ε e−∆ε/τ
[ln(1 + e−∆ε/τ )] = 2
∂τ
τ 1 + e−∆ε/τ
thus we find the entropy to be
σint
"
∆ε
= N ln(1 + e−∆ε/τ ) +
τ
e−∆ε/τ
1 + e−∆ε/τ
and the total entropy is defined as
"
n 5
∆ε
Q
+ + ln(1 + e−∆ε/τ ) +
σtot = N ln
n
2
τ
126
!#
e−∆ε/τ
1 + e−∆ε/τ
!#
d) The pressure is given as
∂Ftot
p=−
∂V
we can see from b that the total free energy is given as
−∆ε/τ
Ftot = Nτ[ln(n) − ln(nQ ) − ln(1 + e
N
−∆ε/τ
) − 1] = Nτ ln
− ln(nQ ) − ln(1 + e
)−1
V
we can see that the derivative in terms of the volume removes every term except the first one, thus
Nτ
∂F
=−
∂V
V
thus the pressure is given as
p=
Nτ
V
e) The heat capacity at constant pressure is given as
∂σint
5
CP = N + τ
2
∂τ P
where the first term comes from the ideal gas solution without any degrees of freedom. Thus we
need to solve
"
!#
−∆ε/τ
∂σint
e
∆ε
∂
ln(1 + e−∆ε/τ ) +
=N
∂τ
∂τ
τ 1 + e−∆ε/τ
which becomes
∂σint
∂τ
=
=
=
=
"
!
!#
e−∆ε/τ
∆ε e−∆ε/τ
e−∆ε/τ
∆ε ∂
∆ε
N 2
−
+
τ 1 + e−∆ε/τ τ2 1 + e−∆ε/τ
τ ∂τ 1 + e−∆ε/τ
!
∆ε ∂
e−∆ε/τ
N
τ ∂τ 1 + e−∆ε/τ
"
#
−∆ε/τ ) ∆ε e−∆ε/τ − e−∆ε/τ ∆ε e−∆ε/τ
∆ε (1 + e
2
2
τ
τ
N
τ
(1 + e−∆ε/τ )2
"
#
(∆ε)2
e−∆ε/τ
N 3
τ
(1 + e−∆ε/τ )2
thus we find that the heat capacity at constant pressure to be given as
"
5 (∆ε)2
CP = N
+ 2
2
τ
Problem # 6 Isentropic relations of ideal gas
127
e−∆ε/τ
(1 + e−∆ε/τ )2
!#
a) Show that the differential changes for an ideal gas in an isentropic process satisfy
dp
dV
+γ
=0
p
V
dτ
dV
+ (γ − 1)
=0
τ
V
dp
γ dτ
+
=0
p
1−γ τ
where γ = CP /CV . These relations apply even if the molecules have internal degrees of freedom.
We know that for an isentropic gas
pV γ = constant
thus we can see that
τγ/(1−γ) p = constant
τV γ−1 = constant
∂
dp
dp
p
(pV γ ) = V γ
+ pγV γ−1 = 0
= −γ
∂V
dV
dV
V
thus we can see tha
dp
dV
+γ
=0
p
V
for the temperature and the volume relationship we can use
dτ
∂
(τV γ−1 ) = V γ−1
+ τ(γ − 1)V γ−2 = 0
∂V
dV
thus
dτ
dV
+ (γ − 1)
=0
τ
V
and finally for the temperature pressure relationship we can use
γ
∂ γ/(1−γ)
dp
γ ( 1−γ
−1)
=0
(τ
p) = τγ/1−γ
+p
τ
∂τ
dτ
1−γ
thus we can see that this simplifies into
γ dτ
dp
+
p
1−γ τ
b) The isentropic and isothermal bulk moduli are defined as
∂p
∂p
Bτ = −V
Bσ = −V
∂V σ
∂V τ
Show that for an ideal gas Bσ = γp; Bτ = p. The velocity of sound in a gas is given as c = (Bσ /ρ)1/2 ;
there is very little heat transfer in a sound wave. For an ideal gas of molecules of mass M we have
p = ρτ/M, so that c = (γτ/M)1/2 . Here ρ is the mass density.
We have shown that
p
dp
= −γ
dV
V
thus the isentropic bulk moduli is given as
Bσ = −V
∂p
∂V
128
σ
= γp
for the isothermal bulk moduli we must consider
pV = Nτ
and we know that for a constant temperature
dp
p
=−
dV
V
dp
V+p=0
dV
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 constant
entropy, not constant temperature. In such equilibrium pV γ is independent of altitude, where γ = CP /CV .
Use the condition of mechanical equilibrium in a uniform gravitational field to
a) Show that dT /dz = constant, where z is the altitude. This quantity, important in meteorology, is called
the dry adiabatic lapse rate. (Do not use the barometric pressure relation that was derived in Chapter
5 for an isothermal atmosphere.)
We know that the condition of mechanical equilibrium in a uniform gravitational field is given as
V d p = −Nmgdz
where the left hand term is for the energy due to the expansion and the right hand term is due the the
gravitational potential. We can see that
Nmg
dp
=−
dz
V
and we also know that
dT
dT d p
=
dz
d p dz
where
thus
1−γ τ
dτ
=−
dp
γ p
dT
1−γ T
=−
dp
γ p
dT
1 − γ T Nmg
=
dz
γ p V
but we know that
V=
thus we find that
NkB T
p
dT
1 − γ mg
=
dz
γ kB
which is constant.
129
b) Estimate dT /dz, in 0 C per km. Take γ = 7/5.
Using our previous result we find, also using γ = 7/5 yields
2 mg
dT
=−
dz
7 kB
we find the molecular mass of air to be 28.97 amu, thus we find
2 28.97 × 1.66 × 10−27kg(9.8ms−2 )
dT
K
C
=−
=
=
−9.75
−9.75
dz
7
km
km
1.38 × 10−23 (m2 kgs2 K−1 )
c) Show that p ∝ ργ , where ρ is the mass density.
For an isentropic process we know that
pV γ = constant
we know that the mass density is defined as
ρ=
M
N
=m
V
V
thus the volume is given as
V=
mN
ρ
which allows us to substitute into the previous expression as
mN γ
= constant
p
ρ
thus we can see that
p=
so
ργ
(mN)γ
p ∝ ργ
If the actual temperature gradient is greater than the isentropic gradient, the atmosphere may be unstable 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 isothermally
and reversibly to twice the initial volume; second, let this be followed by an isentropic expansion from
twice to four times the initial volume.
• Process 1 is isothermal and reversible thus we know that the pressure after expansion is given as
piVi = p f V f
1
p f = pi
2
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
pdV =
V1
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 know
that under isentropic conditions that
∆σ = 0
∆Q = 0
we can find the temperature using the entropy
σ(τ,V ) = N(ln τ3/2 + lnV + constant)
so that the entropy remains constant if
ln τ3/2V = constant
thus
τ3/2V = constant
so for an expansion at constant entropy from V1 to V2 we have
3/2
3/2
τ1 V1 = τ2 V2
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) = NkB T ln(2)
where we know that this can also be expressed as
Q1 = nRT ln(2)
where R is the molar gas constant, thus we find
Q1 = 1 mole × 8.31 J/K mole × 300 K × ln 2 ≈ 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
(6.5)
The final temperature for process two can be found by using Equation 3, as
τ f = τi
or
T f = Ti
V1
V2
2/3
V1
V2
2/3
2/3
1
= 300 K
2
thus the final temperature is
T f = 189 K
c) Suppose the first process is replaced by an irreversible expansion into a vacuum, to a total volume twice
the 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 properties
hold
∆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 heat
is added in the expansion. Because the energy is also unchanged than the temperature is unchanged. The
increase of entropy when the volume goes from V1 to V2 is given by
σ2 − σ1 = N ln(V2 /V1 ) = N ln(2)
where
N = 1 mole
thus the change in entropy is given as
in standard units this is simply
6.02 × 1023 atoms
= 6.02 × 1023 atoms
mole
∆σ = 4.17 × 1023
∆σ = kB × 4.17 × 1023
this is simply
∆σ = 5.75
132
J
K
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 3/2
2π~2
we know that when n > nQ then this is when quantum behaviours become important. We can also write
this as
mτ
< n2/3
2
2π~
or
2π~2 2/3
n = τ0
τ<
m
if τ ≪ τ0 we call this a degenerate gas and when τ ≫ τ0 we call this a classical gas.
133
7.1.1 Metals
Metals contain “free” electrons that are donated by atoms in the lattice. This gives properties of high
eloectric conductivity. Lets consider two kinds of metals. The alkili metas which are Li, Na,K and Ca
each donate 1 s electrons where for N atoms we have N free electrons. The alkili earth Be,Mg,Ca,Sr, and
Ba each donate 2 s electrons. Thus N atoms give 2N free electrons. At room temperature, the mean free
path is ≈ 40 nm or > 100 lattice spacings. At low temperatures, the mean free path ≈ 400, 000 nm = 400
µm, thus we can see that the electrons are very free. We will now derive the Fermi energy
7.1.2 Fermi Energy
Lets take a box that have a set of well defined energy levels. We know that since it is fermions that we are
constrained by the exclusion principle. The lowest energy level is called the fermi energy level εF . We
know that the energy in the box is given by
ε=
~2 π 2 2
~2 π 2 2
2
2
(n
+
n
+
n
)
=
n
y
z
2mL2 x
2mL2
to relate the number of states in energy to the number of states that lie in n and n + dn
1
N(n)dn = 2 × 4πn2 dn = πn2 dn
8
thus we can see that
εF =
~2 2 2
π nF
2mL2
7.1.3 Fermi Gas
the Fermi energy is given as
~2
εF =
2m
3π2 N
V
V
D(ε) =
2π
2m
~2
3/2
the density of states is given as
3/2
ε1/2
the Fermi density of states is given as
D(εF ) =
3N
2 εF
and the energy is given as
3
3
u(0) = NεF = NkB T
5
5
7.1.4 Pressure of the Fermi gas (T = 0)
We recall that the pressure p is given as
∂u(0)
p=−
∂V
134
σ,N
once again we have u(0) can be written as
3 ~2
3
2 2/3
N 5/3V −2/3
(3π )
u(0) = NεF =
2
5 2m
so we simply need to take the derivative to find
"
#
2
5/3
2 3π2 N 2/3 N
3
2
2
~
N
~
p=−
−
=
(3π2 )2/3
5
3
2m
V
5 2m
V
V
where
"
~2
εF =
2m
3π2 N
V
2/3 #
thus the pressure is given as
2 N
p = εF
5 V
or
2
pV = NkB TF
5
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=
8.3 J mole−1 K−1 × 233
RT
=
≈ 105 Pa
V
22.4 × 10−3m3
now what how does this relate to the Fermi pressure? Lets take sodium as our example, thus
V = 23 cm3 = 23 × 10−6 m3 TF = 35, 000 K
thus
2 RTF
≈ 5 × 109 Pa
5 V
thus the pressure of the Fermi gas to the pressure of an ideal gas at STP is given as
p=
TF 22, 400
pF
≈
×
≈ 105
p
T
23
7.1.5 Specific Heat of Fermi Gas
For insulators at room temperature is CV = 3NkB and for metals the result is basically the same.
135
7.1.5.1 Approximate argument for CV
the number of lectrons that are excited is given by an increase in temperature is
∼N
kB T
T
∼N
εF
TF
thus the change in the energy is
∼N
T
kB T
TF
and thus the heat capacity is
∂
CV =
∂T
T2
NkB
TF
∼ NkB
T
T
∼
× classical value ∼ 0.01(NkB )
TF
TF
for sodium at room temperature. The exact result is given as
π2 NkB T
= γT
CV =
2 TF
where γ is given as
γ=
the total heat capacity of a metal
π2 NkB
2 TF
CV = γT + AT 3
where the first term come from the electrons and the second term comes from the phonon contribution. At
room temperature, the phonos dominate and at low temperature, the elctrons dominate.
7.1.6 Paramagnetic Susceptibility
Calssically
Nµ2
Curie’s Law
χ=
kB T
when we increasee B, only electrons within ∼ kB T of εF ca flip spins, thus
Nµ2B T
Nµ2B
χF ∼
=
kB T TF
kB TF
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 −106 gm cm−3 . This was first discovered from Sirius B which was emperically
determined from the pertubations from Sirius B. We find that the mass of Sirius B is MB = 2 × 1033 g and
the radius is given as rB = 2 × 109 cm.This can be acquired from the Planck distribution and Wiens’s law.
From this, we find that the density ρ ∼ 7 × 104 g cm−3 . We believe that all matter is ionized and thus we
136
have Fermi gases. We must consider the protons and electrons as being two distinct samples of Fermi
gases. If we have
ρ ∼ 1 g cm−3 l ∼ 10−10 m
ρ ∼ 106 g cm−3 l ∼ 10−12 m
and
N
∼
V
1036
1028
2/3
1
10−12
3
m−3 ∼ 1036 m−3
and the Fermi energy is
εF ∼ εF
≈ 3 × 105 eV ⇒ TF ≈ 3 × 109 K
where we know that the temperature needed for fusion is T ∼ 107 K. Since T ≪ TF this is a highly
degenerate gas. Lets do this for protons, lets first assume that the nuclie are protons. The mass of the
proton is approximately 2000me. We can see that
εF ∝
1
m
thus for the protons
TF ∼
3 × 109
∼ 1.5 × 106K
2000
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. The radius is given as
r = (1.3 × 10−13cm)A1/3
then
since N p ≈ Nn thus
N
3A
=
≈ 1038 cm−3 ∼ 1044 m−3
V
4π(1.3 × 10−13cm)3 A
N
∼ 5 × 1043 m−3
V
plugging in the numbers yield
εF = 3 × 107 eV ∼ 30 MeV
TF ≈ 4 × 1011 K
7.1.9 Bose-Einstein Condensation in Liquid 4 He
The helium nucleus has 2 neutrons and 2 protons with a spin of 0 and obeys Bose-Einstein statistics.
This was historically discovered when people started measuring the heat capacity. They measured that
something spectacular happened as you lowered the temperature of the helium liquid, when theyt got to
2.18 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 three
main points
• Computer simulations predict the density from known quantum properties. In fact, the measured
density is about three times less. Interactions will be weakened as they are pushed apart. The
zero point fluctuations in the atoms increase their kinetic energy, thus pushing them further apart
(increase their seperation). ρ ≈ 0.1 g/cm−3 .
• Even for T > Tλ , the thermal conductivity and the viscosity are roughly the same as those of a gas
of the same density.
• It is very hard to solidify helium, helium liquid will always remain liquid at atmospheric pressure
down to arbitrarily low temperatures (lowest temperatures possible). This implies weak interactions
between 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 =
3π2 ~2
π 2 ~2 2
2
2
(1
+
1
+
1
)
=
2mL2
2mL2
the first excited state is given as
ε1 =
and the difference is
ε1 − ε0 =
6π2 ~2
2mL2
3π2 ~2
≈ 10−37 J ≈ 10−18 eV ≈ 10−14 K
2mL2
if we were to cool this to 10−16 K (impossible) vertually all of the atoms will be i the ground state. In fact,
at 1 K virtually all of the atoms are in the ground state. The entire behavior of this gas is controlled mainly
by 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
1
f (ε) =
e(ε−µ)/τ − 1
it is a requirment that the chemical potential always be less then the energy (µ < ε). For convinience we
will set
1
ε0 = 0 f (0, T ) = −µ/k T
B −1
e
lets suppose that the temperature T becomes very small. In the limit
T →0
f (0, 0) = N
where N is the number of atoms in the box, so
f (0, 0) = N = lim
1
T →0 e−µ/τ − 1
138
where µ/kB T ≪ 1
we are really concerned with the behavior of the chemical potential rather than the temperature. Given the
above condition allows us to expand the exponetial
kB T
1
≈−
1 − (µ/kB T ) − 1
µ
hence
kB T
N
this tells us that the chemical potential is just below the ground state energy, but very very close to it.
Hence µ < 0, but very close to ε0 = 0.
µ≈−
Example
lets take
T = 1 K N ∼ 1022
thus
µ ≈ −10−45 J
which is very close to the ground state energy. Thus the probability of the number of particles in the first
excited state is as follows
f (ε1 , 1K) =
1
≈
e(ε! −µ)/τ − 1
1K
kB T
≈ −14 ≈ 1014
ε1 − µ 10
K
thus, at 1 K only 1 atom in 108 is in the first excited state. We can finally calculate the occupancy of the
ground state as a function of temperature.
7.1.11 Temperature dependence of f (0, T )
The first thing we need to do is write down the density of states, this is given as
3/2
V
2m
D(ε) = 2
ε1/2 spin 0
4π
~2
once again lets consider a few energy levels, a ground state and higher energy states, where the total
number of particles is given as
N = N0 + Ne
whereN0 ar the atoms in the ground state and Ne are the atoms in the excited state. Thus
N = N0 (T ) +
Z ∞
D(ε) f (ε, T )dε
0
we must now evaluate this, this is equal to
V
+ 2
N = −µ/k T
B −1
4π
e
1
2m
~2
3/2 Z
0
∞
ε1/2
e(ε−µ)/kB T − 1
dε
as before, we can neglect the chemical potential because it is negligable, thus we have
x=
ε
kB T
139
which allows us to write
V
Ne (T ) = 2
4π
2m
~2
3/2
(kB T )
3/2
Z ∞ 1/2
x
0
V
dx
=
ex − 1
4π2
2m
~2
3/2
(kB T )3/2 (1.31π1/2 )
which can be simplified as
Ne = 0.17V
mkB T
~2
3/2
the fraction of excited states is
V
Ne
≈ 0.17
N
N
mkB T
~2
3/2
≈ 2.61
nQ
n
this a very useful simple result
Ne
nQ
≈ 2.61
N
n
wha this shows us is that the number of excited atoms increases as the temperature to the 3/2. This gives
us a prediction of the transition point for helium. Tλ is given by
Ne
=1
N
thus
~2
Tλ =
2m
N
0.17V
where the observed value is Tλ = 2.18 K. If we write
Ne
=
N
and
T
Tλ
2/3
≈3K
3/2
3/2
N0
T
= 1−
N
Tλ
qualitatively this result is correct, but quantitatively it is incorrect. The model is too simple. What we
have seen is that at low temperatures we have a high concentration of atoms in the ground state. Thus we
have 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 of systems that have the same properties.
Helium 3 atoms are fermions but it turns out that at low enough temperatures these helium 3 atoms can
pair together to give bosons and they can also undergo a condensation to the ground state to create a
super fluid (1972). The third example is super conductivity, this is where we have electrons that become
cooper pairs which act like bosons. (~k ↑, −~k ↓), where they have zero momentum and zero spin. This was
understood theoretically in 1957 BCS.
140
7.1.12 Properties of 4He Superfluids: Thermomechanical effects
Lets suppose we have two resevoirs of helium that are connected by a narrow capillary (pipe). It blocks
nromal atoms but it passes “superfluid” atoms. We then cool the helium down and put pressure on one
side 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 has
less superfluid atoms will increase its temperature. Suppose we take the same equipment and increase the
temperature on one side. Initially both resevoirs are at temperature T and we then increase the temperature
by ∆T on one side. This will create an osmotic pressure of superfluid helium. The level on one side falls
while the level on the other side increases to take care of the pressure difference.
7.2 Summary
For the Fermi gas
~2
εF =
2m
V
D(ε) = 2
2π
3π2 N
V
2/3
the density of states is given as
2m
~2
and
D(εF ) =
3/2
ε1/2
3N
2εF
the ground state energy is given as
3
u(0) = NεF
5
also
2
pV = NkB TF
5
the heat capacity is given as
CV =
π2 NkB T
= γT
2 TF
and also
χ∼
µ2B
kB TF
for the Bose gas
T →0
µ=−
and
V
Ne
≈ 0.17
N
N
mkB T
~2
141
kB T
N
3/2
≈ 2.61
nQ
n
7.3 Problems
Problem # 1 Energy of relativistic Fermi gas
For electrons with an energy ε ≫ mc2 , where m is the rest mass of the electron, the energy is given by
ε ≃ pc, where p is the momentum. For electrons in a cube of volume V = L3 the momentum is of the
form (π~/L), multiplied by (n2x + n2y + n2z )1/2 , exactly as for the nonrelativistic limit.
a) Show that in this extreme relativistic limit the Fermi energy of a gas of N electrons is given by
εF = ~πc(3n/π)
1/3
3N
= ~πc
Vπ
1/3
where n = N/V .
Since we know that we are in the relativistic case that the energy is defined as
ε=
π~c
n
L
but we know that the Fermi density is related to the number of particles by
1 4
π
N = 2 × × πn3F = n3F
8 3
3
nF =
3N
π
1/3
The factor of two arises because an electron has two possible spin orientations. The factor of 1/8 arises
because only triplets in the positive ocatant of the sphere in n space are to be counted, thus the Fermi
energy is given by
~πc
3N 1/3
εF =
nF = ~πc 3
L
L π
which simplifies to
3N
εF = ~πc
Vπ
1/3
= ~πc(3n/π)1/3
b) Show that the total energy of the ground state of the gas is
3
U0 = NεF
4
We know that this can be solved by knowing that the total energy of the system in the ground state is given
by
1
U0 = 2 ∑ εn = 2 × × 4π
8
n≤nF
Z nF
0
π2 ~c
n dnεn =
L
2
Z nF
n3 dn =
0
which simplifies into
3
U0 = NεF
4
142
π2 ~c 4
π2 ~c 3
3π~c
nF =
nF nF =
NnF
4L
4L
4L
Problem # 2 Energy, heat capacity, and entropy of degenerate boson gas
Find expressions as a function of temperature in the region τ < τE for the energy, heat capacity, and
entropy of a gas of N noninteracting bosons of spin zero confined to a volume V . Put the defenite integral
in dimensionless form; it need not be evaluated. The calculated heat capacity above and below τE is
shown in Figure 7.12. The difference between the two curves is marked: It is ascribed to the effect of
interactions between the atoms.
We know that the density of modes is given by
V
D(ε) = 2
4π
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 sum
of the occupancies of all orbitals
N = ∑ fn = N0 (τ) + Ne (τ)
n
and the total energy is given by
U = ∑ εn fn = N0 ε0 (τ) + Neε(τ)
n
but we know that the number of atoms in the excited state is given by
Ne (τ) =
thus the total energy is given by
U=
Z ∞
0
Z ∞
0
D(ε) f (ε, τ)dε
D(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
V
U= 2
4π
2m
~2
3/2 Z
∞
1
λ−1 eε/τ − 1
0
ε3/2 dε
where λ = eµ/τ and if we let x = ε/τ we find
V
U= 2
4π
For λ we can see that
2m
~2
3/2
τ5/2
Z ∞
0
1
x3/2 dx
−1
x
λ e −1
λ=1
this is because we can Taylor expand the activity to find
λ ≈ 1+
1
µ
= 1−
τ
N
143
but we have seen that
−µ =
τ
N
N ≫1
The integral can now be written as
∞
e−x 3/2
x
dx
=
∑
1 − e−x
s=1
Z ∞
0
if we let y = sx then
∞ Z ∞
∑
3/2 −sx
x
s=1 0
e
∞
∑s
dx =
Z ∞
0
−5/2
s=1
x3/2 e−sx dx
!Z
∞
0
e−y y3/2 dy
if we now let y = u2 then we find
Z ∞
0
−y 3/2
e y
and we also find that
dy = 2
Z ∞
0
∞
∑ s−5/2
s=1
2
e−u u4 du =
!
3√
π
4
≈ 1.34
thus we find that the total energy is given by
4.02V
U=
16
3/2
2m
π~2
τ5/2
and we know that the heat capacity is given by
5 4.02V
∂U
=
CV =
∂τ
2 16
2m
π~2
3/2
τ3/2
which simplifies into
CV = 0.63V
and the entropy is given by
2m
π~2
∂σ
CV = τ
∂τ
3/2
τ3/2
V
thus we can see that the total entropy is given by
σ=
Z τ
CV
0
τ
dτ = 0.61V
2m
π~2
3/2 Z
τ
1/2
2
dτ = (0.63)V
3
or
σ = 0.42V
2m
π~2
Problem # 3 Fluctuations in a Fermi gas
144
3/2
τ3/2
2m
π~2
3/2
τ3/2
Show for a single orbital of a fermion system that
h(∆N)2i = hNi(1 − hNi)
if hNi is the average number of fermions in that orbital. Notice that the fluctuation vanishes for orbitals
with energies deep enough below the Fermi energy so that hNi = 1. By definition, ∆N = N − hNi
From Equation 5.83 we know that the fluctuations go as
h(∆N)2 i = τ
∂hNi
∂µ
Equation 6.2 gives that the average number of particles is given by
hN(ε)i = 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 thermal
average number of particles in an orbital of energy ε. Thus for a Fermi gas we find that
hNi =
and
d
∂hNi
=
∂µ
dµ
if we let
e(ε−µ)/τ + 1
1
e(ε−µ)/τ + 1
K = e(ε−µ)/τ + 1
thus
1
d
=
dµ
1
K
dK
1
= − e(ε−µ)/τ
dµ
τ
dN dN dK
=
dµ
dK dµ
and since we know that
N=
we find
1
K
dN
1
=− 2
dK
K
∂N 1
e(ε−µ)/τ
=
∂µ
τ (e(ε−µ)/τ + 1)2
thus the fluctuations is given as
e(ε−µ)/τ
(e(ε−µ)/τ + 1)2
to write it in a more suggestive form we can simply do
1
1
e(ε−µ)/τ + 1 − 1
2
1 − (ε−µ)/τ
= (ε−µ)/τ
h(∆N) i = (ε−µ)/τ
(e
+ 1)2
e
+1
e
+1
h(∆N)2 i =
which can also be written as
h(∆N)2i = hNi(1 − hNi)
Problem # 4 Fluctuations in a Bose gas
145
If hNi as in (11) is the average occupancy of a single orbital of a boson system, then from 5.83 show that
h(∆N)2i = hNi(1 + hNi)
thus, if the occupancy is large, with hNi ≫ 1, the fractional fluctuations are of the order unity:
h(∆N)2 i/hNi ≈ 1, so that the actual fluctuations can be enormous. It has been said that “bosons travel in
flocks.”
As before we know that
hNi = f (ε) =
and as before
thus the fluctuation is given as
1
e(ε−µ)/τ − 1
∂N 1
e(ε−µ)/τ
=
∂µ
τ (e(ε−µ)/τ − 1)2
e(ε−µ)/τ
h(∆N) i = (ε−µ)/τ
(e
− 1)2
2
written in a more suggestive way
1
1
e(ε−µ)/τ − 1 + 1
= (ε−µ)/τ
1 + (ε−µ)/τ
h(∆N) i = (ε−µ)/τ
(e
− 1)2
e
−1
e
−1
2
which can also be expressed as
h(∆N)2i = hNi(1 + hNi)
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 Gibbs
free energy
8.1 Heat and Work: Exact and inexact differentials
We know that heat is a transfer of energy by thermal contact, which is usually denoted as dQ. Work is
a transfer of energy by changes of external parameters, i.e volume, electric field or magnetic fields. For
reversible 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 exact
differential. If we start at some point a to some point b it doesnt matter what path we take, the change in
the energy is independent of the path. But for the change in heat or the cahnge in the work are dependent on
the path chosen. The internal energy u is a function of state while the heat and the work are not functions
of 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 form
C(x, y)dx + D(x, y)dy are of the form as before G(x + dx, y + dy) − G(x, y). We want to now take two
examples to illustrate this point.
8.1.1 Example 1
Lets let
x
dG = αdx + β dy = αdx + βx(d ln y)
y
147
where α and β are cosntants. If we now take the integral, via (1,1) to (1,2) to (2,2)
Z f
i
dG = α + 2β(ln2 − ln 1) = α + 2β ln 2
If we now take the integral, via (1,1) to (2,1) to (2,2)
Z f
i
dG = β(ln 2) + α = α + β ln 2
this is an example of an inexact differential. The real problem is in the term of β. Now for an exact
differential
dG α
dy
dF =
= dx + β = d(α ln x + β ln y)
x
x
y
now if we dow the same thing as last time we find If we now take the integral, via (1,1) to (1,2) to (2,2)
Z f
i
dG = α ln 2 + β ln 2 = α ln 2 + β ln 2
If we now take the integral, via (1,1) to (2,1) to (2,2)
Z f
i
dG = α ln 2 + β ln 2 = α ln 2 + β ln 2
where we can see that these are exactly the same. Any type of work can be converted to any other type of
work (in principle) and entropy is conserved. Lets imagine that we are compressing a gas in a cylinder
via some magnetic interaction. If we attach a bar magnet to the piston and then we move a second
magnet to the first which compresses the gas, thus we have used magnetic work to create mechanical
work. Work can be also be completely converted into heat, i.e stirring 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 heat
at a lower temperature. We have some thermal resevoir at some temperature T1 with some other resevoir at
T2 , and we also have heat Q1 and Q2 and we get out the work W . Where Q1, Q2 , and W are all magnitudes.
The entropy is given as
Q2
Q1
σ2 =
σ1 =
τ1
τ2
hence, the work W is simply
τ1 − τ2
τ2
W = Q1 − Q2 = Q1 − Q1 = Q
τ1
τ1
where this is known as the Carnot efficiency
ηC =
τ1 − τ2
W
=
Q
τ1
this is the maximum possible efficiency that an engine can run on. In practice all engines will have
additional losses and this efficiency can never be achieved. A car engine where τ1 = 600 K and τ2 = 300
K, 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 a resevoir at T1 and another at T2 , we also
have Q1 and Q2 and we put in work W . The Carnot efficiency is the following,
γC =
Q2
Q2
=
W
Q1 − Q2
the exact relationships for the entropy still hold, hence
γC =
Q2
τ1
τ 2 Q2 − Q2
=
τ2
τ1 − τ2
this is the maximum efficiency of a refrigerator. Lets suppose that τ1 = 300 K and τ2 = 260 K, we can see
that
260
= 6.5
γC =
40
This is also how air conditioners work, where T2 is the temperature in the room and T1 is the temperature
of the outside. We can also consider a heat pump where T2 is the temperature outside and T1 is the
temperature T2 and we find that the Carnot efficiency is now given by
γC =
τ1
Q1
Q1
=
=
τ2
W
Q1 − τ 1 Q1 τ 1 − τ 2
we can take an example of τ2 = 270 K and τ1 = 300 K, thus the Carnot efficiency is γC = 10. We can see
that a heat pump is very efficient.
8.2.3 Non ideal case for Heat engine
Just to reacp very briefly, we know that
∆σ = −
τ2
Q1 Q2
+
= 0 ⇒ Q2 = Q1
τ1
τ2
τ2
and the efficiency is given as
ηC =
if we have losses then
W
τ1 − τ2
=
Q1
τ1
∆σ > 0
σ2 > σ1
if we now go back we can see that
Q2 > Q1
for eaxmple, lets set
Q2 = Q1
where δ is the inefficiency, thus we can see that
η=
Q1 − Q1
τ2
τ1
τ2
+δ
τ1
τ2
τ1
Q1
149
+δ
< ηC
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 from σlow to σhi and a fixed temperature
τhi . We then do a second expansion which is adiabatic from σhi and τhi to τlow . We then have to do two
different compressions, the first one is isothermal going from σhi to σlow and at τlow , we then do another
adiabatic compression from τlow to τh at σlow . This are the four steps of the Cranot cycle. We know that
around the entire cycle, the change in energy is equal to zero
0=
I
du
this must also be equal to
I
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 for N molecules and gas using
a p and V diagram. The firs thing is to expand the gas isothermally, thus we know that the pressure drops
and the volume increases. In order to do this we have to apply heat Q1 . We then do an adiabatic expansion
where the volume increases more and the pressure drops more. We then do an isothermal compression at
the 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 the
work done by the gas. We want to figure out the work done by the gas in each of these 4 processes
Z V2
Z V2
V2
dV
= Q1
= Nτhi ln
W1→2 =
pdV = Nτhi
V1
V1
V1 V
then we have the work done during the adiabatic expansion
3
W2→3 = [u(τhi ) − u(τlow )] = N(τhi − τlow )
2
next we go from point 3 to point 4, which is an isothermal expansion
V4
= Q2
W3→4 = Nτlow ln
V3
150
and finally we find
3
W4→1 = [u(τlow ) − u(τhi )] = N(τlow − τhi )
2
and the total work is given as
V2
W = Nτhi ln
V1
V4
+ Nτlow ln
V3
The goal of this calculation is to rederive the Carnot efficiency for this process. To find the Carnot
efficiency we must find a relationship among V1 ...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 we
found the following result
τV 2/3 = constant
we now consider where we have processes at constant entropy, this is for 2 going to 3 and we start with
2/3
τH V2
2/3
= τLV3
in other words
V3
=
V2
and now if we look at 4 to 1 we find
2/3
τLV4
thus
V4
=
V1
τH
τL
3/2
2/3
= τH V1
τH
τL
3/2
=
V3
V2
if we now go back to the expression for the work done we find
V2 V3
=
V1 V4
and finally, what we see is the following
V2
ln
V1
V4
= − ln
V3
therefore the work done by the gas is given by
V2
W = N(τH − τL ) ln
V1
we finally find the Carnot efficiency
ηC =
W
N(τH − τL ) ln(V2 /V1 )
=
Q1
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 temperature τ and a piston that will
compress the gas isothermally. For an isolated system, the internal energy is given as
u = u(σ,V )
and
du(σ,V ) = τdσ − pdV
thermodynamic identity
which reduces to
du = d(στ) − pdV
and the free energy is
dF(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,
where dF is a function of state while dW is not. This is why the Helmholtz free energy is such a useful
function.
8.4.2 Isobaric (constant pressure)
Boiling water, chemical reactions are usually at constant pressure. If we impoise a constant pressure on
our 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 will
provide a pressure on the piston from the vapor. This moves the piston by an amount dx which will give
us dV = Ada. We need to supply heat to evaporate the water and expand the volume to accomdidate the
gas. 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 quantity H which is
dH = d(u + pV ) = dQ
where this new quantity H is known as the enthalpy, 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 the
system, i.e a paddle moving the water. So
du = dQ − d(pV ) + dW
we can therefore write this as follows
dW = du + d(pV )
and also
putting this all together gives
where
dQ = τdσ = d(τσ)
constant τ
dW = d(u + pV − τσ) = dG
G = u + pV − τσ = F + pV
where this is known as the Gibbs Free Energy. Thus the work done on the system is simply
dW = dG
8.4.4 Gibbs Free energy for varying N
We know that
recall that
dG = du + pdV +V d p − τdσ − σdτ
du = τdσ − pdV + µdN
this is the thermodynamic identity when N varies. We can now insert this into the previous line to find
dG = −σdτ +V d p + µdN
so we now have this new term that is very useful. We see immideately that we can write down the following
three results
∂G
∂G
∂G
V=
µ=
−σ =
∂τ p,N
∂p τ,N
∂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 potential µ is the Gibbs free energy per particle. This is really what makes
phase 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 another
inexact 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
τ1
τ1 > τ2
for a refrigirator we find
τ2
can be > 1
τ1 − τ2
we then showed for an isothermal process that the work
ηC =
dW = dF
we also talked about isobaric heat and work allowed us to introduce the enthalpy H which is given by
H(σ, p) = u + pV
and the heat of vaporization
dQ = 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 +V d p + µdN
and hence, we find three very useful results
∂G
−σ =
∂τ p,N
V=
∂G
∂p
154
τ,N
µ=
∂G
∂N
τ,p
8.6 Problems
Problem # 1 Photon Carnot Engine
Consider a Carnot engine that uses as the working substance a photon gas. The Carnot cycle can be
represented as
σ
3
isentropic
2
isothermal
isothermal
σh
1
σl
4
isentropic
τl
τh
τ
The cycle consists of two expansion phases (1 → 2 and 2 → 3) and two compression phases (3 → 4 and
4 → 1). There are two isentropic phases and two isothermal phases. The work done is given by the area
of the rectangle created by the solid line and the heat consumed at τh is the area sorrounded by the broken
line. Since we are considering a photon gas as the working substance of the engine we must consider the
energy density of a photon gas which is given as
U=
but we know that
thus at constant volume we have
integrating both sides yield
π2V 4
τ
15~3 c3
(8.1)
τdσ = dUτ + pdV
4π2V τ2
dτ
dσ =
15~3 c3
4 π2V 3
σ=
τ
45 ~3 c3
a) Given τh and τl as well as V1 and V2 , determine V3 and V4 .
155
(8.2)
We know that the entropy is the same at both 1 and 4 and 2 and 3 thus
σ4 = σ1
using Equation 2 we find
σ2 = σ3
V4 τ3l = V1 τ3h
V2 τ3h = V3 τ3l
thus
V4 = V1
τh
τl
3
V3 = V2
τh
τl
3
b) What is the heat Qh taken up and the work done by the gas during the first isothermal expansion? Are
they equal to each other, as for the ideal gas?
We know that
Qh =
thus
Qh =
Z 2
1
Z
dQ =
τdσ = τ(σ2 − σ1 ) =
Z
pdV =
Z
τdσ
4
4π2 4
τh (V2 −V1 ) = ατ4h (V2 −V1 )
3
3
45~ c
45
where
α=
π2
~3 c3
so we find that the heat taken up is
Qh =
4 4
ατ (V2 −V1 )
45 h
and the work is found by using
dW = dU − dQ
thus the work is
W12 = −
Z
2
1
dU −
Z 2
1
τdσ
the minus sign comes because this is the work done on the gas, thus we can solve to find
4 4
ατ (V2 −V1 )
W12 = − U2 −U1 −
45 h
the change in energy is given by Equation 1 as
U2 −U1 =
thus the work is
or
α 4
τ (V2 −V1 )
15 h
4 4
α 4
τ (V2 −V1 ) −
ατ (V2 −V1 )
W12 = −
15 h
45 h
W12 =
1 4
ατ (V2 −V1 )
45 h
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
W41 = −
dU
2
Z 1
dU
4
since dσ = 0, as before the minus sign comes from the fact that work is being done on the gas we find
W23 = −[U3 −U2 ]
1
4
4
= −
α[τ V3 − τhV2 ]
15 l
"
#!
3
1
τ
h
W23 = −
α τ4l V2
− τ4hV2
15
τl
W23 =
1 3
ατ V2 [τh − τl ]
15 h
and also
W41 = −[U1 −U4 ]
1
4
4
= −
α[τ V1 − τhV4 ]
15 l
"
3 #!
τh
1
α τ4l V1 − τ4hV1
W41 = −
15
τl
W41 = −
1 3
ατ V1 [τh − τl ]
15 h
thus
1 3
ατ [τh − τl ][V2 −V1 ]
15 h
we can see that they do not cancel each other out as in the case for the ideal gas.
W23 +W41 =
d) Calculate the total work done by the gas during one cycle. Compare it with the heat taken up τh and
show that the energy conversion efficiency is the Carnot efficiency.
To find the total work done we must finally consider the work done during
Z 4
Z 4
W34 = −
dU −
τdσ = − (U4 −U3 − τl (σ4 − σ3 ))
3
3
which yields
W34 =
=
=
=
α 4
4 4
−
τ (V4 −V3 ) − ατl (V4 −V3 )
15 l
45
α
τ4 (V3 −V4 )
−
45 l
3 !!
3
τh
τh
α 4
τl V2
−V1
−
45
τl
τl
α
− τl τ3h (V2 −V1 )
45
157
where the total energy is given by
W = W12 +W23 +W34 +W41
1 4
1
1
=
ατh (V2 −V1 ) + ατ3h [τh − τl ][V2 −V1 ] − ατl τ3h (V2 −V1 )
45
15
45
1 3
1 3
ατ [τh − τl ][V2 −V1 ] + ατh [τh − τl ][V2 −V1 ]
=
45 h
15
4 3
W =
ατ [τh − τl ][V2 −V1 ]
45 h
and comparing this to the Carnot efficiency yields
τh − τl Th − Tl
W
=
=
Qh
τh
Th
ηC =
Problem # 2 Geothermal Energy
A very large mass M of porous hot rock is to be utilized to generate electricity by injecting water and
utilizing the resulting hot steam to drive a turbine. As a result of heat extraction, the temperature of the
rock drops, according to dQ = −MCdTh , where C is the specific heat of the rock, assume to be
temperature independent. If the plant operates at the Carnot limit, calculate the total amount W of
electrical energy extractable from the rock, if the temperature of the rock was initially Th = Ti , and the
plant is to be shut down when the temperature has dropped to Th = T f . Assume that the lower resevoir
temperature Tl 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−1 K−1 , Ti = 6000 C, T f = 1100 C, Tl = 200 C. 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 =
thus the work is given as
dW =
Th − Tl
W
=
Th
Qh
Th − Tl
Th − Tl
dQ = −
MCdTh
Th
Th
thus the total work is given as
W = −MC
Z Tf
Th − Tl
Ti
Th
Tf
dTh = MC [Ti − T f ] + Tl ln[ ]
Ti
plugging in numbers given by
Ti = 873 K
T f = 383 K
W = 2.48 × 1016
Tl = 293 K
kg J
= 2.48 × 1019 J
g
in kWh this is simply
thus the work done is given by
1 kWh = 3.6 × 106 J
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 by
changing 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 superconducter. Finally normal Helium 4 to superfluid Helium 4.
Road map
1) Gas/liquid/solid phase transformations.
• p vs V isothermal
• p vs T coexistance curves- triple point
2) Clausis-Clapeyon Equation
Latent heat, vp vs T . Change is boiling point with pressure and latent heat of vaporization of ice
3) Van Der Walls Equations
• departures from ideal gas
• predictions of p vs V
• critical points at which all hree phases can exist
9.1 Gas, Liquid, and Solid Phase Transformations
9.1.1
p vs V
If we look at a plot of the pressure versus the volume for a liquid curve we find that at low volumes we
have an isotherm. The reason is due to the incompressability of liquid. We start in a liquid state with a
piston pushing hard against it. As we keep removing the pressure we get to a point where the piston is
raised above the liquid. The pressure than becomes constant, and this is where we can have both liquid
and gas (this all occurs in T < 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 of
these transitions. In liquid stage we find
µl < µg
in the liquid and gas stage we find
µl = µg
and finally in the gas stage
µg < µl
9.1.2 Coexistence Curve p vs T
If we look at a plot of prsseure versus temperature we find that at low temperatures, we have a line that is
very nearly to verticle hat drops down to 0 degrees. Anything on the left we have ice, so that the chemical
potential 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 the
left 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 and Tc ≈ 374 degrees
celsius. We also have no distinction between the liquid and the gas. To recover the previous plot we
simply need to just take a slice in our current plot to get what are known as coexistence curves. The
particularly interesting things happen at the origin where we have have all three phases of water occuring
at the same point. This is known as the triple point of water Pt and Tt . The triple point is actually used to
define the temperature scale. For water
Tt = 273.16 K = 0.01 0 C
and
Pt = 4.58 mm Hg
this is used in thermometry. The line between solid and the liquid is called the fusion curve, we also have
the sublimation curve, that is the line along which gas can be turned directly to a solid, and finally we
have the vaporization curve, that is the line which liquid is transformed into a gas. The fusion curve is
called the ice line, the sublimation curve is called the frost line and the vaporization curve is caled the
steam 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 has Po and T0 and the second
has P0 + dP and T0 + dT . We can say the following that the chemical potential of these two points must
be 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
∂µg
∂µl
∂µl
µg (P0 , τ0 ) +
dP +
dτ = µl (P0 , τ0 ) +
dP +
dτ
∂P τ
∂τ P
∂P τ
∂τ P
we can see that the first terms go away, we can now collect terms to get
∂µg
∂µl
∂τ P − ∂τ P
dP
= ∂µ g
dτ
l
− ∂µ
∂P
∂P
τ
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 also
dG = µdN − σdτ +V dP
as before
V=
and we can see that
∂G
∂P
∂µ
V =N
∂P
τ,N
τ,N
∂G
σ=−
∂τ
σ = −N
∂µ
∂τ
N,P
N,P
we can now define the volume per molecule and entropy per molecule
∂µ
V
v= =
N
∂P τ,N
this is the volume per molecule, and similarly
σ
∂µ
s= =−
N
∂τ N,P
which is the entropy per molecule. We can now substitute these results into our initial result. We find
dP sg − sl
=
dτ
vg − vl
now we have the pressure and temperature in a simple form. The numerator
sg − sl
corresponds to the increase in the entropy when one molecule changes from a liquid to a gas. And correspondingly
vg − vl
161
corresponds to the increase in the volume when one molecule changes from a liquid to a gas. We can now
look at the latent heat of vaporization L, this is the heat required to transfomr one molecule from the
liquid to the gas phase. We can relate this to τdσ
L = τ(sg − sl ) = τ∆s
or
we thus find
and
dQ = τdσ
∆v = vg − vl
L
dP
=
dτ τ∆v
this is known as the Clausius-Clapeyron Equation. This confirms the results from the Coexistence
curves. We can also write this in various forms that are approximations.
9.1.3.1 Approximate results
We must assume that vg ≫ vl , an example, at 1 atm vg ≈ 1000vl . Therefore we can write
∆v = vg =
vg
τ
=
Ng P
this assumes the ideal gas law. Hence
L
LP
dP
=
= 2
dτ τ∆v
τ
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
Z
dP
dτ
=L
P
τ2
if we take water for example, at 0 degrees celsius, the latent heat is 2520 J/g and at 100 degrees celsius the
latent heat is 2260 J/g. We can see that this is approximately 12 percent change. We now see that
L
ln P = − + ln P0
τ
and finally
P = P0 e−L/τ
this confirms the exponential behaviour of our coexistence curve. It is quite common to see this expression
written as
L0 = NA L
(NA is Avegadros number)
gives us
P = P0 e−L0 /RT
we can easily plot this function. The above expression is not completely correct but should be expressed
as
L L
ln P − ln P0 = − +
τ τ0
thus
−L
P = P0 e
162
1− 1
τ τ0
9.1.4 Examples
9.1.4.1 Calculate dT /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
thus
L0 P
dP
=
L0 = 2260 J/g × 18 = 4.07 × 104 J/mole
2
dT
RT
also R = 8.31 (J/K mole)thus we can see that
dT
RT 2
=
= 28.4 K/atoms
dP
L0 P
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, find L of ice at -1 degrees centigrade. We will use our result
L0 P
dP
=
dT
RT 2
thus
L0 =
lets consider
RT 2 dP
d ln P
= −R
P dT
d(1/T )
d ln P dT
d ln P
=
d(1/T )
dT d(1/T )
we can write the following
1
d(1/T )
=− 2
dT
T
thus
d ln P
d ln P −T 2 dP
= −T 2
=
d(1/T )
dT
P dT
we just need to plug in numbers into
L0 = −R
0.7
d ln P
= −R
= 5.1 × 104 J/mole
d(1/T )
−2.7 × 10−5
163
9.1.5 Latent Heat and Enthalpy
If we know the specific heat of something at constant pressure we 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 generally
we can write the Gibbs free energy as
G = ∑ N jµ j
j
we can then write dow the thermodynamic identity
du = τdσ − PdV + ∑ N j dµ j
j
now we want to talk about the transformation from a liquid to a gas. Some number of molecules will
change from being a liquid to a gas
dN = dNg = −dNl
this is all valid in the coexistence curve. Therefore
du = τdσ − PdV + (µg dNg + µl dNl )
we can see that the last term vanishes in the coexistance state, thus
du = τdσ − PdV
therefore
L = τdσ = du + PdV
but we know that the enthalpy is defined as
dH = d(u + PV ) = du + PdV +V dP
but we know the pressure is constant, thus
L = dH = Hg − Hl
and also
∂σ
Cp = τ
∂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 the
volume correction first
9.2.1 Volume correction
We can reduce the effective volume to a value
V ′ = V − Nb
where V is the volume of the box and b is some other volume. The concentration is changed by
N
N
n′ = ′ =
V
V − Nb
in practice b 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 called the mean
field value.This refers to the idea that we neglect fluctuations. Lets suppose that we have a quantity φ(r)
which is the potential energy of the interaction between the two atoms seperated by a distance r. We have
a concentration n, the number of atoms per unti volume. The average potential energy of all other atoms
relative to one particular atom is written as
average value of all other atoms on one atom at position r = 0
=
Z ∞
b
φ(r)ndV
we have assume that n is a constant value. The mean value “field” implies that all the atoms or molecules
are always at their “mean” position, that is to say we neglect fluctuations (using n′ will give only a second
order correction). We can write the integral as
−2na =
thus
Z ∞
b
Z ∞
b
φ(r)ndV
φ(r)ndV = −2a
it is minus because it is an attractive intercation, attractive force. The factor of 2 is purely for convention
and 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 =
N
(−2na)
2
where the 1/2 comes from double counting. Thus
dF = −
N2
a
V
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
N2
nQ (V − Nb)
+1 − a
= −Nτ[ln(nQ /n ) + 1] = −Nτ ln
N
V
′
and the pressure is
∂FdW
p=−
∂V
τ,N
nQ
N2
N
Nτ
N2
= − −Nτ
+ 2a =
− 2a
nQ (V − Nb) N
V
V − Nb V
165
we finally get the following
N2
p + 2 a (V − Nb) = Nτ
V
this is the van der Waals equation for a gas. If you think of the interactions of atoms or molecules you find
that at very low seperations there is a repulsive force which is usually modeled as ≈ 1/r12 . On the other
hand if you start pulling the hydrogen molecule apart there will be an attraction given by 1/r6 , which is
the van der Waals term. Where the 1/r6 can be looked as the 1/V 2 term arising from the van der Waals
force. We can now use this equation to predict critical behavior.
9.2.3 Finding critical point
If we look at the p versus V plot we can see that we have three critical points for phase transitions. There
are a series of isotherms that can also be seen in the plot. Aobe some critical point we get what is called
a fluid and this happens at the critical isotherm which occurs at τC . These isotherms occur on the vapor
staturation curve. The thing that will enable us to find the critical point is that
2 ∂p
∂ p
=0
=0
∂V τC
∂V 2 τC
knowing that
thus
and
N2
Nτ
− a
p=
V − Nb V 2
∂p
∂V
=−
Nτ
N2
+
2
a=0
(V − Nb)2
V3
⇒
2Na
τ
=
V3
(V − Nb)2
2Nτ
N2
3Na
τ
∂2 p
=
−
+
6
a=0 ⇒ 4 =
2
3
4
∂V
(V − Nb)
V
V
(V − Nb)3
If we devide the equations we find
2VC
= VC − Nb
3
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 and b). If we take the previous expression
and insert into the first equation we find
τC =
2Na
8 a
2
(2Nb)
=
27N 3 b3
27 b
(9.2)
we also want to find pC from the van der Waals equation
pC =
4Na 1
N2a
4 a
3 a
a
− 2 2=
−
=
2
2
27b 2Nb 9N b
27 b
27 b
27b2
(9.3)
hence VC and τC define a and b. We want to eliminate the three parameters a, b and N from the van der
Waals equations. If we start with
N2
p + 2 a (V − Nb) = Nτ
V
166
we can write
VC
p+
3b
2
27b2 pC
V2
!
VC
V−
3
=
=
8V C τpC
3 τC
VC
τ
3b
we still need to get rid of b, we can devide 9.3 with 9.2 we get
a 27b
1
pC
=
=
2
τC
27b 8a
8b
thus we find
VC
p+
3b
2
27b2 pC
V2
!
VC
V−
3
this is an equation that involves p,V and τ and pC ,VC and τC . We can divide through by the product pCVC
thus
2 ! VC
V
1
8 τ
p
+3
−
=
pC
V
VC 3
3 τC
where we are only left with the ratios of all these quantities. We can now introduce
p̂ =
p
pC
V̂ =
V
VC
τ̂ =
τ
τC
which are rthe reduced values, thus we can write
3
p̂ +
V̂
8
1
= τ̂
V̂ −
3
3
two gases with the same values of p̂ V̂ and τ̂ are said to be in corresponding states, i.e all gases look the
same in terms of these three variables. Lets go back to the previous expression and write it as
1
8
( p̂V̂ + 3) V̂ −
= τ̂V̂ 2
3
3
this is a general cubic expression in V̂ . This expression can be solved numerically, but in general a cubic
function gives an S shaped curve with two turning points. If you decrease the pressure and increase the
volume we come to a metastable state, and likewise if you decrease the volume and increase the pressure
also allow you to reach a metastable state. The turning point for the first case is called superheating and
the second case is supercooling. This formalism breaks down when V̂ < 1/3. If we consider the line for
pA (lower left minimum) we can write the Gibbs freen energy as
dG = −σdτ +V d p + µdN
but at constant τ and N we find
dG = V d p
so as we go from the left minimum to the right maximum then the total change from this is given as
∆G =
Z B
dG = 0
A
167
we can write down the integral in the following way
Z D Z C Z E Z B Z B
Vdp =
+
+
Vdp
+
A
A
D
C
E
Z C Z A Z E Z E Vdp
−
=
−
−
E
D
B
C
where the first two integrals is the area of ADC and the second set of integrals is the area of CBE and this
is all equal to 0. We define p̂ 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 do you build liquid drops from gases. We are
talking about metastability. 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 surface
tension (energy) increases the chemical potential for the liquid and may prevent condensation and this
leads to the concept of the critical radius of the drop. If the drop is above this radius it can grow but if it
is below this area than it shrinks
R > Rc
R < Rc
grows
shrinks
recall that the Gibbs free energy is given as
G = Nµ
and the change in the Gibbs free energy is given as
4
∆G = Gl − Gg = πR3 nl (µl − µg ) + 4πR2 γ
3
where γ = G/A is the Gibbs free energy per unit area. The first term dominates for large radius and the
second term dominates for small radius. To find the critical radius we take the derivative
d∆G
= −4πR2 nl ∆µ + 8πRγ = 0
dR
thus the critical radius is given as
Rc =
2γ
nl ∆µ
if we take the second derivative
d 2 ∆G
= −8πRnl ∆µ − 8πγ = −16πγ + 8πγ = −8πγ < 1
dR2
thus this is a maximum. What is the height of the barrier?
4
∆Gb = − πR3c nl ∆µ + 4πR2c γ
3
168
substituting our expression for the critical radius we find
4γ2
4 8γ3
∆G = − π 3 3 nl ∆µ + 4π 2 2 γ
3 nl ∆µ
nl ∆µ
3
πγ
32
=
16 −
3 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 sorrounded
by vapor which applies a pressure p > pbulk and p(τ) exceeds pbulk (τ), where the bulk case µl = µg . Recall
that
n
= τ ln(Ap)
µ = τ ln
nQ
since we know that p ∝ n at fixed τ. Thus
∆µ = µ p − µ p(bulk) = τ ln(Ap) − τ ln(Apbulk ) = τ ln
p
pbulk
we have related the change in chemical potential to the ratio of these two pressures. Putting some numbers
for water
τ = 300 K
p
pbulk
thus
Rc =
= 1.1
nl =
6 × 1023 1023 3 1029 3
=
cm =
m γ = 7.2 × 10−2J m−2
18
3
3
2γ
14.4 × 10−2
m = 10−8 m
= 29 10
nl ∆µ
−23
(1.4 × 10 × 300 ln(1.1))
3
9.4 Summary
The Clausius-Claperyin equation
L
dp
=
dτ τ∆V
where
L=
Lt · ht
molecule
and the pressure is given as
−
P = P0 e
and
L0 =
L0
R
1
1
T − T0
Lt · ht
mole
and the van der Waals equation is given as
N2
p + 2 a (V − Nb) = Nτ
V
169
the critical point is given as
Vc = 3Nb
pc =
a
27b2
τc =
8a
27b
and corresponding states
3
p̂ +
V̂
1
8
V̂ −
= τ̂
3
3
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
nQ (V − Nb)
5
σ = N ln
+
N
2
We know that the free energy of the gas, with the van der Waals correction is given as
N2
nQ (V − Nb)
+1 − a
FdW = −Nτ ln
N
V
and we know that the entropy is defined as
∂F
σ=−
∂τ
V
"
"
!
#
#
∂
N2
kτ3/2 (V − Nb)
=−
−Nτ ln
+1 − a
∂τ
N
V
where k contains all of the constants that come with the quantum concentration. We know that
∂A
∂B
∂
(AB) = B + A
∂τ
∂τ
∂τ
if we let
A = −Nτ
B=
3
ln(kτ) + ln(V − Nb) − ln N + 1
2
thus we find
3
∂ 3
∂
(AB) = −N
ln(kτ) + ln(V − Nb) − ln N + 1 − Nτ
ln(kτ)
∂τ
2
∂τ 2
3
3
ln(kτ) + ln(V − Nb) − ln N + 1 +
= −N
2
2
nQ (V − Nb)
5
= −N ln
+
N
2
thus the entropy is given as
nQ (V − Nb)
5
+
σ = N ln
N
2
b) Show that the energy is
3
a
U = Nτ − N 2
2
V
170
We know that the total energy is given as
U = F + τσ
thus
N2
5
nQ (V − Nb)
nQ (V − Nb)
+ 1 − a + Nτ ln
+
U = −Nτ ln
N
V
N
2
3
a
=
Nτ − N 2
2
V
thus the total energy is given as
N2
3
U = Nτ − a
2
V
c) Show that the enthalpy H = U + pV is
bτ
a
5
Nτ + N 2 − 2N 2
2
V
V
5
ap
H(τ, p) =
Nτ + Nbp − 2N
2
τ
H(τ,V ) =
All results are given to first order in the van der Waals correctio terms a, b.
If we start with
N2
p + 2 a (V − Nb) = Nτ
V
expanding this equation gives
pV − pNb +
N2
N3
a − 2 ab = Nτ
V
V
thus
N3
N2
a + 2 ab + Nτ
V
V
since all the results are given to first order allows us to simplify
pV = pNb −
pV = pNb −
N2
a + Nτ
V
thus the enthalpy is terms of the volume gives
5
bτ
a
H(τ,V ) = U + pV = Nτ + N 2 − 2N 2
2
V
V
and since we know that p = Nτ/V ⇒ V = Nτ/p and also N/V = p/τ, thus the enthalpy in terms of the
pressure give
5
ap
H(τ, p) = Nτ + Nbp − 2N
2
τ
Problem # 2 Latent heat of magnesium
171
In the temperature range 700 to 730 K, the vapor pressure of magnesium can be approximately represented
by
7, 500
+ 8.6
log10 p = −
T
(p in mmHg, T in K). What is the molar latent heat of sublimation of magnesium in this temperature
range?
We know that the latent heat of is given as
L = τ(s1 − s2 )
where s1 and s2 are the entropy for the specific phase of the subtance, thus 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 also know that
ln p = −
L0
+ constant
τ
but we are given
7, 500
+ 8.6
T
so it is just a matter of comparing these two equation and doing a change of base. We also know that
log10 p = −
ln p = loge p =
thus
log10 p
log e
log10 p
= ln p
log e
and we find
log10 p = −
7, 500
8.6
+
T log e log e
and we can see that the latent heat is given by
L0 =
J
J
7, 500kB
= 2.38 × 10−19 = 1.43 × 105
log e
K
mole
and so
L0 = 1.43 × 105
Problem # 3 Latent heat of neon
172
J
mole
The latent heat of fusion and the latent heat of vaporization of neon at the triple point have values
335J/mole and 1810 J/mole respectively. The temperature of the triple point is 24.57 K. What is the latent
heat of sublimation of neon at the triple point? What is the change in entropy when one mole of liquid
neon 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
Lf
τ
thus we can see that the latent heat of sublimation is
ss = sl −
sg =
Lv
+ sl
τ
Ls = Lv + L f
thus we find that the laten heat of sublimation is
Ls = 2145
J
mole
and the change in entropy for vaporization is given by
sg − sl =
Lv
Lv
=
× mole
τ
kB T
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 the
notion of quantized energy in a box. We can combine this with the probability that any one of these states
is occupied, i.e combine quantum energy levels with f (εl ), recall
n −εl /τ
f (εl ) =
e
nQ
~2
εl =
2m
where
πl
L
2
we will confine our system to have only spin 0 atoms. We want to express the number of atoms with
quantum numbers between l and l + dl this goes as
1
π n
N(εl ) = 1 × × 4πl 2 × dl × f (εl ) = l 2 e−εl /τ dl
8
2 nQ
where 1 is for the spin 0 and 81 4π2 dl
tum numbers (l) to speed (v), thus
number of states in shell of octant we want to convert from quan~2
1
εl = mv2 =
2
2m
πl
L
2
this is the important step that takes you from quantum numbers into speed
ML
~2 π 2 l 2
l=
v
2
2
m L
π~
if we say that there are N molecules in a volume V and lets introduce a probability that a particle has a
velocity in a particular rangep(v) is the probability that molecules have speeds between v and v + dv, thus
v2 =
1
n
dl
N p(v)dv = πl 2 e−εl /τ dv
2 nQ
dv
174
where
ML
dl
=
dv
π~
putting this together we have
1 n
N p(v)dv = π
2 nQ
ML
π~
2
2 −Mv2 /2τ
v e
lets recall
n
N
= 3
nQ L
2π~2
Mτ
ML
dv
π~
3/2
and putting everythign together we find
πN
N p(v)dv =
2 L3
2π~2
Mτ
3/2 ML
π~
3
2 /2τ
v2 e−Mv
dv
we can see that a lot of terms cancel
M
p(v) = 4π
2πτ
3/2
2 /2τ
v2 e−Mv
this is the famous Maxwell distribution for speeds of a classical ideal gas. We can now derive the RMS
speed for the distribution.
Three speed
The RMS speed is defined as follows
2
hv i =
Z ∞
0
where we can write
M
4π
2πτ
3/2
thus
4
hv i = 2
π
2
we can set y =
Mv2 /2τ
3/2 Z
4
= 2
π
M
v p(v)dv = 4π
2πτ
2
M
2τ
3/2 Z
0
∞
0
∞
M
2τ
2 /2τ
e−Mv
3/2
2 /2τ
e−Mv
v4 dv
thus
v=
2τ
M
1/2
1/2
dv =
y
and
1
v dv =
2
4
2τ
M
5/2
175
2τ
M
1/2
y3/2 dy
1 −1/2
y
2
v4 dv
and finally we find
Z
M 3/2 2τ 5/2 ∞ −y 3/2
hv i = 1/2
e y dy
2τ
M
π
0
3 2τ 3τ
=
=
2M
M
2
4
therefore
r
3τ
M
this is the same result we derived in the equipartition theorem, i.e
1 2 3
mv = τ
2
2
Lets now calculate the mean speed which is not zero, but we know that the maen velocity is zero. We can
write
Z
Z ∞
4 M 3/2 ∞ −Mv2 /2τ 3
vp(v)dv = 2
hvi =
e
v dv
π 2τ
0
0
making the same substitution as before gives
1 2τ 2
3
ydy
v dv =
2 M
plugging this in gives
3/2 2 Z ∞
2τ
4
M
e−y ydy
hvi = 1/2
2τ
M
π
0
which yields
8τ 1/2
hvi =
πM
q
hv2 i =
thus this is still of the form τ/M, thus the maen speed is slightly different than the RMS speed. We can
now derive what the most probable speed. We can see that at low speeds, the distribution is quadratic, but
with increasing speed the exponential term will dominate. The maximum will give us the most probable
speed. The most probable speed can be derived by differentiating the previous expression
2 /2τ
p(v) = Av2 e−Mv
where
thus
Mv3 −Mv/2τ
d p(v)
e
= A 2v −
dv
τ
v2MP =
2τ
M
and finally
r
2τ
M
this are the three speeds that gases can have. We can normalize to find
v
v
=p
vMP
2τ/M
vMP =
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 atom between collisions. It is almost always
given as l. We can replace the fact that atoms of like kind have a diameter d which are all moving around in
different directions. We replace the previous picture by assuming that one atom has a diameter 2d (super
atom) and all other atoms are points. We can consider our super atom will move a distance l in a cylinder
with length l. On the average there is one atom in the cylinder of volume V = lπd 2 , that is to say since
there is one atom in the cylinder
lπd 2 = n = 1
where n is the number of atoms per unit volume, thus
l=
1
πnd 2
how big is this number?
10.2.1 Estimate of l
Lets first assume that we are in STP conditions. We also have the Loschmidt number
n=
Avegadros number
6.02 × 1023
=
= 2.69 × 1025 m−3
molar volume at 0 celsius and atm. pressure 22.4 × 10−3m3
we can now make an estimate of the mean free path
l=
1
π × 2.69 × 1025 m−3 × 2 × 10−10 m2
≈ 300 nm
how often does this collision occure? The mean free time is found by
t=
3 × 10−7 m
l
=
≈ 0.3 ns
v
103 m/s
lets take another limit, for example an ultra high vacuum system
p ∼ 10−13 atm
thus
l ∼ 3 × 10−7 m × 1013 ∼ 3000 km
the mean free path becomes really large under low pressures. This forces very different behaviors for
gases under different conditions. The mean free time in this system 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 as transport. We are assuming
that 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 a
higher 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 concentration µl > µr . The flux of particles is
defined by Fick’s law, which says
dn
Jnz = −D
dz
where D is the diffusive constant (diffusivity), the minus sign comes from the fact that the gradient is
positive to the left but particles flow to the right. The flow is from high n to low n. We have defined lz as
the mean free path along the z axis and vz is the z component of the average velocity. At the plane z we
have the flux in the +z direction is given by
1
n(z − lz)v¯z
2
and the flux in the -z component is
1
n(z + lz)v¯z
2
and the net flux is the difference between these two components
Jnz =
1
[n(z − lz) − n(z + lz)] v¯z
2
we can write the term on the left, with a Taylor expansion
n(z + lz) ≈ n(z) + lz
dn
dz
and likewise for the other term, thus
dn
v¯z lz
dz
we have not taken into account for the fact that not all these atoms travel in the z direction, thus we must
take an angular average to account for this. Thus the projection on the z axis is given by
Jnz = −
vz = v̄ cos θ
and
lz = l cos θ
where lz is the projection of the MFP on the z axis. Recall that the solid angle is defined as
dΩ = 2π sin θdθ
we want
hvz lz i =
v̄l2π
R π/2
0
2π
cos2 θ sin θdθ
R π/2
0
sin θdθ
where the denominator is for normalization. Thus we can see
"
#π/2
R π/2
− 13 cos3 θ
v̄l 0 cos2 θ sin θdθ
v̄l
hvz lz i =
= v̄l
=
R π/2
− cos θ
3
sin θdθ
0
0
178
thus the diffusion constant is
D=
v̄l
3
how big is this constant? Lets take a gas at STP. Recall that v̄ ≈ 105 cm/s and l ∼ 3 × 10−5 cm and thus
D ∼ 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 concentration gradient with one that has a temperature
gradient. Imagine that we have two reseviors with temperature τ1 and τ2 where τ1 > τ2 . Thus we now
have Fourier’s law
∂τ
Jµz = −K
∂z
where K is known as the thermal conductivity. We want to write this as follows
Juz = −D
∂ρu
∂z
where ρu is the energy density, thus this is an exact analogy from what we wrote before, we can write
Juz = −D
where we can see that
∂ρu ∂τ
∂τ ∂z
∂ρu
= CV
∂τ
thus
Juz = −DCV
∂τ
∂z
and hence
K = DCV =
v̄l
CV
3
this result shows us that the thermal conductivity is independent of the pressure of the gas, CV ∝ n and
l ∝ 1/n and hence K is independent of n and hence the pressure. This is provided that the mean free path
is still much less then the dimensions of the box. How big is it? At STP, the heat capacity is
3
3
CV = kB n ∼ × 3 × 1019 cm−3 × 1.4 × 10−23 J/K ∼ 10−3 J/K cm−3
2
2
and thus
K = CV D ∼ 10−3 J/Kcm3 × 1 cm2 /s ∼ 1 mW/cmK
179