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8/19/2019 States With Given FIXED ENERGY Are ACCESSIBLE States-112_formulas

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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

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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 asystem with given energy which is equal to the number of quantum states with that energy ε. e.g Particlein 3D box

n x n y n z1 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 2 N , the total magnetic moment is given by M = Nm for allspins that are ↑ and M = − Nm for all spins that are ↓. How many ways can we get M = ( N −2)m (flipover 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

2 + s

↑ n↓ =

N

2 − s

↓

where N is even and M = (n↑ − n↓)m = 2sm where the 2s = spin excess = n↑ − n↓. If we assume eachspin has a probability P↑ of being ↑ and a probability of P↓ of being ↓ . we also know that P↑ + P↓ = 1. Inthe presence of a magnetic field ↑, P↑ > P↓. The probability of a configuartion with spin excess 2s isP

N /2+s↑ and P

N /2−s↓ . The probability of finding a magnetic moment M = (2s)m is given by the following

P( N ,s) = g( N ,s)P N /2+s↑ P

N /2−s↓

we need to find what the multiplicity g( N ,s) is. In how many ways can we distribute n↑ = N /2 + s spinsover 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.

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• 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) sidesThus 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

2 + s

yields

g( N ,s) = N !

N 2 − s! N

2 + s

!

This is the number of ways of achieving a spin excess of 2s. Hence

P( N ,s) = N !

N 2 − s! N

2 + s

!

P N /2+s↑ P

N /2−s↓

For the moment lets assume that we dont apply any field B = 0then P↑ = P↓ = 12 , thus we find

P( N ,s) = N !

N 2 − s! N 2

+ s!1

2 N

If we have the following spins

N spins ↑ ↑ ↓ ↑ ↑ ↓the probability is given as

P( N ,s) = g( N ,s)

2 N 2s = spin excess

where

g( N ,s) = N !

N 2 − s! N

2 + s

!

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1.2 Gaussian Distribution

we first need to assume

N ≫ 1, N even |s| ≪ N P↑ = P↓ = 12

we will use ln as the natural logrythim, thus

ln g( N ,s) = ln N ! − ln

N

2 + s

! − ln

N

2 − s

!

working on this expression yields N

2 + s

! =

1,2,3...

N

2

N

2 + 1

,...

N

2 + s

=

N

2

!

N

2+ 1

,...

N

2+ s

ln N 2

+ s! = ln N 2! +

s

∑k =1

ln N 2

+ k and now we need to do it for the other one

N

2 − s

=

1,2,3, ..

N 2 −s N

2 −s + 1 ,... N

2 N 2....

N 2 − s + 1

=

N 2

!

N /2......

N 2 −s + 1

ln

N

2− s

! = ln

N

2

!0 −

s

∑k =1

ln

N

2−k + 1

hence

ln

N

2 + s

! + ln

N

2 − s

! = 2 ln

N

2

! +

s

∑k =1

ln

1 + 2k N

1 − 2k N +

2 N

and since we know that s ≪ N and also k ≪ N , we can expand the expressions

∑k =1

ln

1 +

2k

N

− ln

1 − 2k

N +

2

N

≈

s

∑k =1

2k

N −

−2k N

+ 2

N

≈ 4 N

s

∑k =1k −

1

2when we add all these terms together and find that there are s terms, thus the sum is given by

4

N

s

∑k =1

k − 1

2

=

2s2

N

putting this all together yields

ln g( N ,s) = ln N ! −2 ln

N

2

! − 2s

2

N

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thus

ln

g( N ,s)

N 2

!2

N !

= −2s

2

N

and

g( N ,s) = N !

N 2 ! N 2 ! e−2s2/ N

and to summarize

g( N ,s) = g( N ,0)e−2s2/ N where g( N ,0) =

N ! N 2

!

N 2

!

we also know that this function falls of as 1/e when

2s2

N = 1, s =

N

2

1/2s

N =

1

2 N

1/2

thus, if N is very large, the peeking is exceedingly sharp. For large N , thermodynamic quantities (heatcapacity,paramagnetic susceptibility are very well defined. To take an example, lets assume that

N = 1020 (1 cm3air) s

N ≈ 10−11

How to evaluate N !? There is a useful approximation called Sterling’s approximation, it is defined as

for large N N ! ≈ (2π N )1/2 N N e− N

we need to work with this until we make it look like the multiplicity g( N ,s)

ln N ! = N ln N − N + 12

ln(2π N )

note that the error is less the 1% for N = 10 and it gets much better for higher N . Lets use this to evaluateg( N ,0)

ln g( N ,0) = ln

N !

N 2

!

N 2

!

= N ln N − N + 1

2 ln(2π N ) −2

N

2 ln

N

2 − N

2 +

1

2 ln(π N )

let us do

ln g( N ,0) = N ln 2 + 1

2ln(2π N )− 1

2ln(π N )2 = N ln 2 +

1

2ln

2

π N

or more compactly

ln g( N ,0) = ln

2 N

2

π N

1/2

and finally

g( N ,0) = 2 N

2

π N

1/2

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1.2.1 Notes on The Gaussian

g( x,σ) = Ae−( x− ¯ x)2/2σ2

where

¯ x = mean value

= 0 for P↑ = P↓= 0 for P↑ = P↓σ = 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 ValuesSuppose that function f (s) has a probability of P(s) of occuring. Then the mean value is

f = ¯ f = ∑s

f (s)P(s) where ∑s

p(s) = 1

e.g, the toss of a die

n̄ =6

∑n=1

np(n) = (1 + 2 + 3 + 4 + 5 + 6) × 16

= 31

2

Example: Spin system

for P↑ = P↓ (i.e equally likely for spins to be ↑ or ↓)thus

P( N ,s) = g( N ,s)

2 N

and using the definition

f = ¯ f = ∑s

f (s)g( N ,s)

2 N

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if we let

f (s) = s2

then

s2 = s2but

s2 = ∑s

s22 N

2π N

1/2e−2s

2

/ N 12 N

=

2

π N

1/2 Z ∞−∞

s2e−2s2/ N ds

if we let

x2 = 2s2

N s =

N

2

1/2 x ds =

N

2

1/2dx

and so we find

s2 = 2

π N 1/2

N

2 N

2 1/2 Z ∞

−∞

x2e− x2

dx = N

4

now lets calculate the spin excess

(2s)2 = N mean square(2s)21/2 =

√ N root mean square

to find the fractional rms spin excess

(2s)2 N

= 1√

N

1.5 Summary

1. Multiplicity for N spins 1/2 with spin excess 2s is given by

g( N ,s) = N !

N 2 + s

!

N 2 − s!

but for N ≫ 1, and s ≪ N we can writeg( N , .s) = g( N ,0)e−2s

2/ N

where

g( N ,0) = N !

N 2 !

N 2 !

= 2 N

2

π N

1/2

and using Sterlings approximation

ln N ! = N ln N − N + 12

ln(2π N )

2. the Mean value

f = ∑s

f (s)P(s), where ∑s

P(s) = 1

and the spin system

(2s)2 N

= 1√

N

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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

π N

1/22 N e−2s

2/ N

And the probability is given as

P( N ,s) = g( N ,s)2 N

and since we know that

N = 400 s = x − ¯ x = 215 − 200 = 15we find that the probability of getting 215 heads is given as

P(400,15) =

2

π(400)

1/2e−2(15)

2/400 ≈ 1.3%

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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 ¯ x = x = N p = 200030

. We find that this distribution can be modeled as a Gaussian distribution of

the form

The distribution plotted above is given by

p( x) = 1√

2πσ2e−( x− ¯ x)

2/2σ2

whereσ2 = 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


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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↑ = N

2 + s n↓ =

N

2 − s

and if we let

n↑ = N

2 + s = N s =

N

2

and we also know that the probability is given as

P( N ,s) = g( N ,s)P N /2+s↑ P

N /2−s↓

where the probability of the number of lines being used is given by

P↑ = 2 min

60 min

2000 lines

N lines =

200

3 N

and the other probability is given as

P

↓ = 1

−P

↑ =

3 N −200

3 N The Gaussian approximation for the multiplicity is given as

g( N ,s) = 2 N

2

π N

1/2e−2s

2/ N

but since we know that s = N 2 we get

g( N , N /2) = 2 N

2

π N

1/2e− N /2

thus we find that the probability is given by

P( N ,s) = 2 N

2

π N

1/2e−2s

2/ N

200

3 N

N /2+s3 N −200

3 N

N /2−s(1.1)

given that s = N 2

this simplifies into

P( N , N /2) = 2 N

2

π N

1/2e− N /2

200

3 N

N

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and since we know that the probability needs to be 0.01% we can just write this as

2 N

2

π N

1/2e− N /2

200

3 N

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 !

( N − n)!n! pn(1 − p) 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

p(n


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since we know that

σ =

N p(1 − p) ¯ x = 200030

thus we find

x =√

2σ(1.822) + ¯ x

thus we find that in order to get 99% we need

x = 86.681 = 87 lines

we have just basically shown that this problem can be done 4 different ways all yielding similar result,

it is just a matter of the approximation that we make.

Problem # 3

The binomial distribution can be written in the form

P( N ,n) =

N

n

pn(1 − p) 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

P( N ,n) = λne−λ

n!

where λ = N p. This is the well-known Poisson distribution (first derived to study the death rate fromhorse kicks in the French army!). Sketch P( N ,n) vs. n for λ 1.

Since we know that

P( N ,n) = N

n pn(1 − p) N −n

where the binomial coefficient is N

n

=

N !

( N −n)!n!or using Newton’s generalized binomial theorem

N

n

=

N ( N − 1)( N − 2)...( N − (n −1))n!

but since we know that N ≫ n this is simply

N n

= N

n

n!

thus we can write our original expression as

P( N ,n) = ( N p)n

n! (1 − p) N −n

we can see that that

(1 − p) N −n =∞

∑k =0

N −n

k

(− p)k

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but we can see that N −n

k

=

( N −n)( N − n −1)( N − n − 2)...( N −n − (k −1))..k !

= N k

k !

since we know that N ≫ n. Thus we find

(1 − p) N −n =∞

∑k =0

(− N p)k

k ! = e−( N p)

thus we find

P( N ,n) = ( N p)n

n! e−( N p)

but if we let N p = λ then we recover what we were looking for

P( N ,n) = λn

n!e−λ

and the Poisson curves are given by

We can see from the plots that the Poisson distribution is well defined for large values 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 havetyped out a single specified book in the age of the universe?

Suppose that 1010 monkeys have been seated at typewriters throughout the age of the universe, 10 18 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 H amlet ?

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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 H amlet ) 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

p(1) = 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 ) = 100000log10

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 approxi-mately 10−164316. The probability of H amlet 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

N monkeys = 1010 t universe = 10

18 s N keys/s = 10 keys

s monkey

thus the total number of keys that these monkeys can type in the total age of the universe is

N keys = 1010 monkeys× 1018 s × 10 keys

s monkey = 1029 keys

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

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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 thistwo-state system is given by

P( N ,s) = g( N ,s)

2 N

we also know thatg( N ,s) =

N ! N 2

+ s

!

N 2 − s!

thus

g(10,0) = 10!

5!5!

and the probability is given as

P(10,0) = 10!

5!5!

1

210 = 0.25

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 maketeh Sterling approximation

g( N ,s) = g( N ,0)e−2s2/ N

where

g( N ,0) = 2 N

2

π N

1/2and so the probability is given by

P( N ,s) = g( N ,s)

2 N =

2

π N

thus we find

P(1000,0) =

2

1000 × π = 0.025

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.

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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, n x,

n y,

and n z

.

• One particle would remain in its state indefenitelyNow we add more particles- ecah one of these is in a state specified by these quantum numbers (n x,n y,n z)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.

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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) = N !

n↑!n↓!=

4!

3!1!= 4

Thus there are 4 possible states, so lets now take four systems, where each one is accesible to system

↓ ↑ ↑ ↑

↑ ↓ ↑ ↑↑ ↑ ↓ ↑↑ ↑ ↑ ↓

This is our ensemble, lets suppose that the probability of finding the system i an accesible state k with

spin excess 2 is

2 = p(k )

where

∑k

p(k ) = 1

Lets suppose that we have a parameter x that takes the value x(k ) when the system is i the state k , thiscould be the magnetic suceptibility or the heat capacity. Then the expectation value of x is

x = ∑k

p(k ) x(k ) ensemble average

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 com-

pare 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 ) = 1

g

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the total energy

u = −2smB = −2(s1 + s2)mB = −2(s′1 + s′2)mBWe also need to assume that N 1


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Lets plug some numbers in to evaluate this, lets take a cm3 of a solid, the number of atoms in this solid is

N 1 = N 2 = 1022

lets suppose that δ is

δ = 1012 s

N 1= 10−10

Thus2s2

N = 200

i.eg1g2

(g1g2)max= e−400 ≈ 10−173

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 (g1g2)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

10−12 = 1034/sec

so the time taken to get this deviation of

δ

N 1= 10−10 ∼ 10

173

1034 ∼ 10139sec ∼ 10132yrs

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

g1g2 = g1(0)g2(0) exp

−2s

2

N 1− 2(s − s1)

2

N 2

if we take the log of both sides we get

ln g1g2 = ln[g1(0)g2(0)−2s2

N 1 −2(s

−s1)

2

N 2

if we take the derivative we find

∂

∂s1[ln g1g2] = −4s1

N 1+

4(s − s1) N 2

= 0 ⇒ s1 N 1

= s − s1

N 2=

s2

N 2

thus the fractional spin excess are equal, if we take the second derivative we find

∂2

∂s21[lng1g2] = − 4

N 1− 4

N 2


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so atŝ1

N 1=

ŝ2

N 2=

s

N

which is the fractional excess of the combined system, thus we have

ŝ1

N 1=

ŝ2

N 2=

s − ŝ1 N 2

= s − ŝ1 N

− N 1

which becomes

ŝ1

1

N 1+

1

N − N 1

=

s

N − N 1thus

ŝ1 =

N 1 − N 1 + N 1 N 1( N − N 1)

=

s

N − N 1which simplifies into

ŝ1

N 1=

s

N

using this result we find

ŝ12

N 1+

ŝ22

N 2=

ŝ1 ŝ1

N 1+

ŝ2 ŝ2

N 2=

s

N ( ŝ1 + ŝ2) =

s2

N

thus we can conclude that

(g1g2)max = g1(0)g2(0)e−2s2/ N

2.5 Thermal Equilibrium: Entropy and Temperature

We have the expression for the multiplicity of the two systems given as

g( N ,u) =

∑u1≤ug

1( N

1,u

1)g

2( N

2,u

2)

if we assume that we keep only the most probable configuration, we can drop the sum to get

g( N ,u) = g1( N 1,u1)g2( N 2,u2) MPC

if we take the derivative we find

dg( N ,u) =

∂g1∂u1

g2du1 +

∂g2∂u2

g1du2 = 0

where

du1 + du2 = 0if we now take this expression and devide

1

g1

∂g1∂u1

N 1

= 1

g2

∂g2∂u2

N 2

this relates the derivative of the first system to the second system, hence∂ ln g1

∂u1

N 1

=

∂ ln g2

∂u2

N 2

22


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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 = k B ln g

as a final remark, Entropy is additive

σ( N ,u) = ln[g1( N 1,u1)g2( N 2,u2)

= ln g1( N 1,u1) + ln g2( N 2,u2)

= σ1( N 1,u1) + σ2( N 2,u2)

We can define the temperature as1

τ =

∂σ

∂u

N

thus in equilibrium

τ1 = τ2

conventionally τ is written asτ = k BT

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 = ∑u1

g1( N 1,u1)g2( N 2,u2) = (g1g2)max

we also assumed in our discussion of entropy that

σ = ln g( N ,u) = ln(g1g2)max = σ1 + σ2

the correct expression is really

σ = ln∑g1( N 1,u1)g2( N ,u −u1)

= σ1 + σ2Why can we neglect all the other terms? How valid is our approximation? We have the following expres-

sion

g1( N 1,s1)g2( N 2,s2) = (g1g2)max

exp−2δ2

N 1 −2δ2

N 2 δ = s1 − ŝ1

lets suppose that we write

N 1 = N 2 = N

2

we get

g( N ,s) = ∑δ

g1( N /2, ŝ1 + δ)g2( N /2, ŝ1 −δ)

= (g1g2)max

Z ∞−∞

e−8δ2/ N d δ

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so in our analysis we have just ignored the gaussian, the integral is

Z ∞−∞

e−8δ/ N d δ =

N

8

Z ∞−∞

e− x2

dx =

N π

8

hence

g( N ,s) = (g1g2)max N π

8

next we can use Sterling’s approximation

g( N ,0) = 2 N

2

π N

1/2 N → N

2

we get the following

g( N /2,0) =

2 N /2

4

π N

2 N π

8= 2 N

4

π N

N π

8

finally we need to form the entropy

σ( N ,s) = ln g( N ,s) = N ln 2 + ln

4

π N

+

1

2ln

N π

8

we can rewrite this as follows

σ( N ,s) = N ln 2 − ln N + 12

ln N + 1

2 ln(π/8) + ln(4/π)

= N ln 2 − 12

ln N

if we take a big number N = 1022

N ln 2 ∼ 1022 ln N ∼ 2.3log10 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 configu-

ration 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 thermalequilibrium

τ1 = τ3

τ2 = τ3

τ1 = τ2

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• 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 = ∑u g1(u1)g2(u −u1)somewhere in this sum is the term that corresponds to the starting point. What we will call that as

g(u) = gi1(u1(t = 0))gi2(u −u1(t = 0)) + ∑

u1=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(gi1g

i2) + ln

∑g1g2

where

σi = ln(gi

1gi

2)

we know that the final value of the entropy is

σ f ≥ σiIf we can imagine having two systems where τ1 > τ2 and that δu is removed from system and added tosystem 2. What happens to the entropy when we do this

∆σ =

δσ

δu1

N 1

(−δu) +

δσ

δu2

N 2

(δu)

∆σ = δu 1

τ2 −

1

τ1≥

0

when τ1 > τ2 then δu > 0

• The Third Law of Thermodynamics- The entropy of a system approaches a constant value as thetemperature goes to zero.

This will follow from our statistical approach if temperature is an increasing function of energyδT

δu

N

> 0

again we will have two systems where we have δu removed from system 2

δu1 = δu δu2 = −δu τ1 = τ2 = τagain we have to go back to our law of increase of entropy, what happens to ∆σ1 and ∆σ2, this are thevalues at t = 0. We can do this by using a Taylor expansion

σ1 = σ0 +

∂σ1∂u1

δu

u0

+ 1

2

∂2σ1

∂u21

u0

(δu)2 + ..

σ2 = σ0 +

∂σ2∂u2

(−δu)

u0

+ 1

2

∂2σ2

∂u22

u0

(−δu)2 + ..

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and

∆σ1 = δu

τ +

(δu)2

2

∂

∂u

∂σ1∂u1

u0

∆σ2 = −δuτ

+ (δu)2

2

∂

∂u

∂σ2∂u2

u0

and finally

∆σ = ∆σ1 + ∆σ2

is∂

∂u

∂σ

∂u

=

∂

∂u

1

τ

= − 1

τ2∂τ

∂u

thus

∆σ = (δu)2

2

− 1

τ2∂τ1∂u

− 1τ2

∂τ2∂u

= −(δu)

2

2τ2

∂τ1∂u

+ ∂τ2∂u


27/179

• We then defined the temperature as1

τ =

∂σ

∂u

N

τ = k BT

• 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 = τ3 τ2 = τ3 τ1 = τ2

1st = heat = energy

2nd = ∆σ ≥ 03rd = lim

τ→0σ = σ0


Problem # 1 Entropy and Temperature

Suppose g(U ) = CU 3 N

/2

, where C is a constant and N is the number of particles.

a) Show that

U = 3

2 N τ

Since we know

g(U ) = CU 3 N /2

we can take the natural log of both sides

ln g(U ) = lnCU 3 N /2 = ln(C ) + 3 N 2

ln(U )

Taking the partial derivative on both sides with respect to U yields

∂

∂U ln g(U ) =

∂

∂U

ln(C ) +

3 N

2 ln(U )

=

3 N

2

∂

∂U ln(U )

we know that the left hand term is simply the entropy

σ = ln g(U )

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thus∂σ

∂U =

3 N

2

1

U

thus

U = 3 N

2

∂U

∂σ

but we know that the temperature is defined as

1

τ =

∂σ

∂U

N ,V

thus we have just showed that

U = 3 N

2 τ

b) Show that

∂2σ

∂U 2 N


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with σ0 = log g( N ,0). Further, show that 1/τ = −U /m2 B2 N , where U denotes U , the thermal averageenergy.

Since we know that

σ0 = log g( N ,0) U = −2smBthus

s = − U 2mB

thus Equation 1 becomes Equation 2, i.e

σ(U ) = σ0 − U 2

2m2 B2 N

if we take a differential on both sides with respect to U we find

∂σ(U )

∂U = − U

m2 B2 N

but we know that1

τ =

∂σ

∂U = − U

m2 B2 N

so we find that the average thermal energy is given by

U = −m2 B2 N

τ

this is related to the spin excess and the magnetic moment by

s

=

−

U

2mB

= mBN

2τwe find that the equilibrium value at temperature τ of the fractional magnetization is given as

M

Nm=

mB

τ

Problem # 3 Addition of entropy for two spin systems

Given two systems of N 1 ≃ N 2 = 1022 spins with multiplicity functions g1( N 1,s1) and g2( N 2,s−s1), theproduct g1g2 as a function of s1 is relatively sharply peaked at s1 = ŝ1. For s1 = ŝ1 + 10

12, the product g1g2is reduced by 10−174 from its peak value. Use the Gaussian approximation to the multiplicity function;the form

g1( N 1, ŝ1 + δ)g2( N 2, ŝ2 −δ) = (g1g2)maxe− 2δ2

N 1− 2δ2

N 2

(2.3)

may be useful.

a) Compute g1g2/(g1g2)max for s1 = ŝ1 + 1011 and s = 0.

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For s1 = ŝ1 + δ we know that δ = 1011

δ = 1011 N 1 ≃ N 2 = 1022 s1 + s2 = s ⇒ s1 = −s2We can see that Equation 3 can be written as

g1( N 1,s1)g2( N 2,s

−s1)

(g1g2)max = e−4δ2/ N 1

≈ 0.018For s = 0 we know that δ = 0 thus we know that

g1g2

(g1g2)max= 1

b) For s = 1020, by what factor must you multiply (g1g2)max to make it equal to ∑s1 g1( N 1,s1)g2( N 2,s −s1); give the factor to the nearest order of magnitude.

Method 1

since we know that

g1( N 1,s1)g2( N 2,s2) = g1( N 1, ŝ1 + δ)g2( N 2, ŝ2 − δ) = (g1g2)maxe− 2δ2 N 1 −

2δ2

N 2

since

s1 = ŝ1 + δ s2 = ŝ2 −δthus

∑s1

g1( N 1,s1)g2( N 2,s − s1) = (g1g2)max ∑δ

e

− 2δ2

N 1− 2δ2

N 2

= (g1g2)max × Q

where Q is the factor we need to multiply (g1

g2)max so that it is equal to ∑s1

g1( N 1,s1)g2( N 2,s−

s1). Sowe can just turn this sum into an integral

Q =Z ∞−∞

e

− 2δ2

N 1− 2δ2

N 2

d δ =

Z ∞−∞

e

− 4δ2

N 1

d δ

this is simply a Gaussian integral with a solution

Q =

π N 1

4= 8.86 ×1010 ≈ 1011

Method 2

We know that

Q × (g1g2)max = ∑g1( N 1,s1)g2( N 2,s − s1) (2.4)but we know that we can write the sum as

∑g1( N 1,s1)g2( N 2,s − s1) = g( N ,s)

since we know that

N = N 1 + N 2 = 2 N 1

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thus we find

g(2 N 1,s) = g(2 N 1,0)e−2s2/2 N 1 = g(2 N ,0)e−s

2/ N 1

we also know that

(g1g2)max = g1( N 1,0)g2( N 2,0)e−2s2/2 N 1 = g1( N 1,0)g2( N 2,0)e−s

2/ N 1

thus Equation 4 can be written as

Q = ∑ g1( N 1,s1)g2( N 2,s − s1)

(g1g2)max=

g(2 N 1,0)

g1( N 1,0)g2( N 2,0) (2.5)

but we know that

g( N ,0) =

2

π N 2 N

and Equation 5 becomes

Q = 22π N 1 22 N 1 2π N 1

2π N 1

22 N 1 = π N 1

4 = 8.86 ×1010

≈ 1011

c) How large is the fractional error in the entropy when you ignore this factor?

We know that the fractional error is defined as

∆σ

σ =

σ′ − σσ′

where

σ′ = ln(g1g2)maxQ σ = ln(g1g2)maxthus we find

∆σ

σ =

lnQ

ln(g1g2)maxQ=

ln Q

ln Q + ln(g1g2)max=

ln Q

ln Q + ln g1( N 1,0)g2( N 2,0) − s2 N 1but we know that

(g1g2)max = g1( N 1,0)g2( N 2,0)e−s2/ N 1

and we also know that

g1( N 1,0) = 2

π N 1 2 N 1

g1( N 1,0) = 2

π N 2 2 N 2

since we know that N 1 = N 2 = 1022, plugging this into some calculating machine we find

∆σ

σ ≈ 1.83 × 10−21

this shows us that the extra factor does not affect the entropy very much.

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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 temper-

ature τ and also that the energy of the system is εl ≪ u0. We must consider an entire ensemble of suchsystems 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) = g R(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) = g R(u0 − εl)and since this total system is equally likely to be in any of of these states thus the probability is

pl = Ag R(u0 −εl )

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and since this is a probability

∑l

pl = 1

again we have an approximation which implies that we will have to do a Taylor expansion, but we will

find that these multiplicity functions will begin to get out of hand when Taylor expanded. We will instead

Taylor expand the entropy

σ = ln[g R(u0 − εl)] ≈ ln[g R(u0)] −

εl∂ ln[g R]

∂u

u0

+ ...

= ln

g R(u0 −εl )

g R(u0)

= −εl

τ

If we raise both sides to the exponetial

g R(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 ≪ U 0where l is to denote the state of the system. What we are looking for is the ensemble average X

X = ∑l

X (l) p(l)

we need to find the multiplicity function of the combined system

g(U 0) = g R(U 0 −εl)gs(l)

where

gs(l) = 1

by definition. Thus the probability we are looking for is proportional to g(U 0), and to do this we mustTaylor expand to find

g(U 0 − 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

εn =

n +

1

2

ω

we can see that the higher the energy the lower the probability of finding this system in this state

pn+1

pn=

e−(n+3/2) ω/τ

e−(n+1/2) ω/τ = e− ω/τ

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and we can see

τ → 0 pn+1 pn

→ 0

τ → pn+1 pn

→ 1

and we also know that

∑l

pl = ∑l

Ae−εl/τ ⇒ A

∑l

e−εl/τ−1

where

pl = e−εl/τ

∑l e−εl/τ

where the denominator is known as the partition function

Z ≡ ∑l

e−εl/τ

this is useful if you start with a microscopic model where we know how to calculate the microstateenergies. From these microstate energies we can use the partition function to get the thermodynamic

properties of this system. If we define the energy states then we can use the partition function as

Z n = ∑n

gs(εn)e−εn/τ

where gs(εn) is how many states have ε = εn. Lets find the average energy given the partition function

U ≡ ε = ε̄where

ε

= ∑

l

εl p(l) = 1

Z

∑l

εle−εl/τ

where we often use β = 1/τ, using this we get

ε = ∑l

εl p(l) = 1

Z ∑

l

εl e−εl β

and the trick we can use∂

∂βe−εl β = −εl e−εl β

and now we can write

ε = 1 Z −

∂ Z

∂β = −∂ ln Z

∂β

to put this back in terms of τ we just simply need to use the chain rule

∂ Z

∂τ =

∂ Z

∂β

∂β

∂τ =

∂ Z

∂β

− 1

τ2

thus

ε = 1 Z

τ2∂ Z

∂τ = τ2

∂ ln Z

∂τWe will need to calculate what the scale of the energy fluctuations are, we will use a method that was

used in quantum mechanic.

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3.2 Energy Fluctuations

(∆ε)2 mean squared energy fluctuationwhere

(ε −ε)and

(∆ε)2 = ε2 −2εε + ε2= ε2− 2εε + ε2= ε2−ε2

we can use the same trick we have done before

ε2 = 1 Z

∑l

ε2l e−βεl

where the trick is

∂2∂β2

e−βεl = ε2l e−βεl

thus

ε2 = 1 Z

∂2 Z

∂β2

∂

∂β

1

Z

∂ Z

∂β

= − 1

Z 2

∂ Z

∂β

2+

1

Z

∂2 Z

∂β2

we can write this as

− 1 Z 2

∂ Z

∂β

2+

1

Z

∂2 Z

∂β2 = −

1

Z

∂ Z

∂β

2+ ε2 = −ε2 + ε2

to finish up(∆ε)2 = ∂

∂β

∂ 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 m(τ). The system consist of a single spin and the resevoirhas all the other spins with respect to τ. We will also neglect all the spin-spin interactions.

m

(τ

) = 1

Z ∑lm

le−

βεl

what are the states? there are only two states, 1 toward the field and one opposing the field

m↑ = +m ε↑ = −mB spin upm↓ = −m 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

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so the expectation value is now

m(τ) = meβmB −me−βmB

eβmB + e−βmB = m

sinh(βmB)

cos(βmB) = m tanh(βmB)

thus

m(τ)

= m tanh(βmB)

but we are looking for the moment for all N spins

M (τ) = N m(τ) = Nm tanh

βm

τ

we can also find the suceptibility to be

χ = d M

dB=

Nm2

τ sech2

mB

τ

if we look at the limits of the hyporbolic tangent function

tanh( x) = e x − e− xe x + e− x

thus for small x we find

lim x→0

tanh( x) = (1 + x) − (1 − x)

1 + x + 1 − x = 2 x

2 = x

and for large x we find

lim x→∞ tanh( x) ≈ 1

In the limit that mB/τ ≪ 1 or high temperature

M (τ) ≈ Nm

Bm

τ

=

Nm2 B

τ

and

χ = Nm2

τ Curie’s law

in the limit of low temperature we find

M (τ) = Nmand

χ = 0

another thing we are interested in is to find the average energy U = ε

ε = − M (τ) B = − NmB tanh

mB

τ

thus the heat capacity is defined as

C V =

∂U

∂T

V

k B

∂U

∂τ

=

∂U

∂T

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where∂U

∂τ = − NmBsech2

mB

τ

−mB

τ2

thus the heat capacity is

C V = k B Nm

2 B2

τ2 sech2

mB

τ

lets look at what the heat capacity does in the extremes. At high temperatures

sech( x) = 2

e x + e− x =

2

1 + x + 1 − x = 1and

C V = Nm2 B2

τ2

and at low temperatures

sech( x) = 2

e x = 2e− x

in this case the heat capacity is roughly

C V = Nm2 B

2

τ2

2e−mB/τ

2 = 4 Nm

2 B

2

τ2 e−2mB/τ

if we look at this graphically we can see that there is a maximum that is known as the Schotkki anomoly.

If you have interaction we get another phenomenom

χ → 1τ − τC

where τC is known as the Curie temperature. This form tells us that when the temperature is lower thanthe Curie temperature forces a negative susceptibility and we have a feromagnetic material and not para-

magnetic.

At high temperatures the magnetic spins are all equally divided in up states and down states. At low

temperatures all of the spins are in the up (high energy) state and the magnetic energy is higher than thethermal energy. If all the spins are aligned with he field it will lower the energy and if all the spins are

anti-aligned that raises the energy.

3.4 Calculating the average energy using the partition function

we can write the average energy as

ε− 1 Z

∂ Z

∂β = −∂ ln Z

∂β = τ2

∂ ln Z

∂ Z

where we know that the partition function is given as Z = emβ B + e−mBβ = 2 cosh(mβ B)

and thus∂ Z

∂β = 2mB sinh(mβ B)

so we have

ε = − 1 Z

∂ Z

∂β = −2mB sinh(mβ B)

2cosh(mβ B) = −mB tanh(mβ B)

which is the exact result we have already obtained.

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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

N 2

N 1 =

e− E 2/τ

e− E 1/τ = e−( E 2

− E 1)/τ

> 1

where E 2 − E 1 is assumed positive, for this to be true means that τ


39/179

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 ap proximation 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 (V + ∆V ) = ε(V ) + dV

∂εl∂V

σ

where the lower case sigma means that this process is at constant entropy. The ensemble average is given

by

U (V + ∆V ) = U (V ) + dV

∂U

∂V

σ

= U (V ) − pdV and thus

p = −∂U

∂V σwe need to know how to get pressure in terms of entropy. What we know is that we can write

σ = σ(U ,V , N ) where N is constant

If we let

0 = d σ =

∂σ

∂U

V , N

(dU )σ +

∂σ

∂V

U , N

(dV )σ

and now if we do

0 =

∂σ

∂U

V , N

∂U

∂V

σ

+

∂σ

∂V

U , N

where we know that1

τ = ∂σ

∂U

V , N − p = ∂U ∂V σ

and we finally find

p

τ =

∂σ

∂V

U , N

3.6 Thermodynamic Identity

To simplify this we will write σ = σ(U ,V ) and we will keep N fixed, thus

d σ = ∂σ

∂U

V dU +

∂σ∂V

U

dV

= 1

τdU +

p

τdV

we can now write this as

τd σ = dU + pdV

thus we can write the change in the energy as

dU = τd σ − pdV where τd σ is the heat added to the system and − pdV is the work done on the system.

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3.7 Helmholtz Free Energy

We have used the free energy in the thermodynamic identity as

dU (σ,V ) = τd σ− pdV

this does not always have to be the case, imagine a systme at constant temperature. We could write this

instead asdU = d (τσ) − σd τ − pdV

where

d (U −τσ) = −σd τ − pdV and

dF (τ,V ) = −σd τ − 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

dF (τ,V ) =

∂F

∂τ

V

d τ +

∂F

∂V

τ

dV

and hence

σ = −

∂F

∂τ

V

p = −

∂F

∂V

τ

where F is the appropriate energy for isothermal systems (fixed τ). If we look at the definition for thepressure

p =

−∂F

∂V

τ

=

−∂(U −τσ)

∂V

τ

=

−∂U

∂V

τ

+ τ∂σ∂V

τ

where the first term is the energy part and the second term is the entropy part. The first tirm tends 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

σ = −∂F ∂τ V p = −∂F

∂V

τ

If we differentiate ∂σ

∂V

τ

= −

∂2F

∂V ∂τ

∂ p

∂τ

V

= −

∂2F

∂τ∂V

thus we can see

∂σ

∂V

τ

=

∂ p

∂τ

V

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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 −τσimagine a resevoir where

dF s = dU s −τσs fixed temperature

we also have the following

1

τ =

∂σ

∂U

V

dU s = τσs constant volume

if we substitute this into the above equation we find

dF s = 0

is this a maximum or a minimum? Lets consider the total energy of the system

U = U R +U s

σ = σ R + σs = σ R(U −U s) + σs(U s)if we assume U s ≪ U R then we can simply do a Taylor expansion

σ = σ R −U s

∂σ R∂U R

V

+ σs(U s)

we can recall that1

τ =

∂σ R∂U R

V

putting this in gives

σ = σ R(U ) − 1τ

(U s −σsτ)we can see that

σ = σ R − F sτ

this is because that we are assuming that σ R(U ) is a fixed quantity. We know that in thermal equilibriumthat the entropy is a maximum, therefor, if we consider a small deviation from σ

∂σ = ∂σ R − ∂F sτ

= −∂F sτ

thus for τ > 0 then if σ is a maximum means that F s is a minimum. On the other hand the energy of thesystem corresponds to a minimum.

Example: Paramagetism

Lets find the Helmholtz free energy F and the average magnetization M of N spins at τ ≫ mB (hightemperature). We can write the free energy as

F (s,τ, B) = U (s,τ, B) −σ(s,τ, B)τ

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where we know that

U (s,τ, B) = −2smBand also

g( N ,s) = g( N ,0)e−2s2/ N

this is the Gaussian approximation in the high temperature limit. Lets put all this together

F (s,τ, B) = −2smB −τ ln g( N ,0) + 2s2τ

N

finally, in thermal equilibrium we can take the derivative∂F

∂s

τ, B

= −2mB + 4sτ N

= 0

thus

2s = mBN

τ

where

M = 2sm = m2 BN

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 = U − τσ σ = −

∂F

∂τ

V

thus

F = U + τ

∂F

∂τ

V

and we also know

τ2 ∂

∂τ

F

τ

V

= −F + τ

∂F

∂τ

V

and we find

F = U + τ2 ∂

∂τ

F

τ

V

+ F

and finally

U = −τ2 ∂∂τ

F

τ

V

but recall that

U = τ2∂ ln Z

∂τ

and thus we findF

τ = − ln Z

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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

p(εl) = e−εl/τ

Z = e(F −εl )/τ

thus we are able to calculate the Boltzman factor if we know the free energy of the system without ever

knowing what the partition function is.

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

εm =

2

2m π

L2

(n2 x + n2

y + n2

z ) n2 = n2 x + n

2 y + n

2 z n

2 x + n

2 y + n

2 z ≥ 1

we can now write the partition function

Z 1 = ∑n x

∑n y

∑n z

e− 2π2

2mL2τ(n2 x +n

2 y +n

2 z )

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

α2 =

2π2

2mL2τ

thus we can write

Z 1 =Z ∞

0dn x

Z ∞0

dn y

Z ∞0

dn ze−α2(n2 x +n2 y +n2 z )

but this can also be written as

Z 1 =

Z ∞0

e−α2n2 x dn x

3the solution is

Z 1 = π3/2

8

23/2 M 3/2 L3τ3/2

3π3 = V nQ

where nQ is known as the quantum concentration and defined as

nQ =

M τ

2π 2

3/2

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3.11 Average energy

U = τ2

∂ ln Z 1∂τ

V

now

ln Z 1 = 3

2

ln τ + τ independent terms

thus ∂ ln Z 1

∂τ

V

= 3

2

1

τ

or

U = τ2 3

2τ =

3

2τ =

3

2k BT

and the specific heat is given by

C V = 3

2k B

this allows us to measure the mean square velocity

1

2 M v2 = 3

2k BT

thus

v2 = 3k BT M

We want to know what the meaning of the concentration is? Recall the Debroigle wavelength

λ D = h

p ≈ h

M (3k BT / M )1/2 ≈ h

(3 M τ)1/2

and

λ3 D ≈ h3

(3 M τ)3/2 ≈ 1

nQ

thus

nQ ∼ 1λ3 D

this means that the concentration corresponds to one particle in a volume of λ3 D. This allows us to under-stand degeneracy in gases. If 2 particles are separated by a distance much greater then λ D then their wavefunctions 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 cruciallyon whether they are bosons or fermions.

3.11.1 Example: N particles in a box

Imagine that we have a number of particles N 2 at standard pressure and temperature STP. we know that

n ≈ 6 × 1023 molecules/mole

22.4 × 103 cm3/mole ≈ 3 ×1019 molecules/cm3

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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

U = 3

2τ

and that “each” “square” term in momentum and position coordinates has a mean energy τ/2 (k BT /2) inthe classical limit. For an ideal gas, the hamiltonian is given by

H = p2 x

2m+

p2 y

2m+

p2 z

2m

and

U = 3

2τ

Lets imagine that we only have one dimension. Thus the Hamiltonian is given by

H = p2

2m +

1

2kx2

and we have an energy given by

U = τ

Proof

Lets imagine a one dimensional gas with a Hamiltonian

H =

p2

2m = Bp2

i

which is just a general expression for a quadratic term. The energy is given by

ε̄i =

R ∞−∞ εie

−βεi d piR ∞0 e

−βεi d pi=

− ∂∂β

R ∞−∞ e

−βεi d piR ∞−∞ e−βεi d pi

this is simplified as

ε̄i = − ∂∂β

ln

Z ∞−∞

εie−βεi d pi

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and now Z ∞−∞

εie−βεi d pi =

Z ∞−∞

εie−β Bp2i d pi

if we let

y = β1/2 pi

we get Z ∞−∞

εie−βεi d pi = 1 β

Z ∞−∞

εie− By2 dy

but we know that

ln

1

β

Z ∞−∞

εie− By2 dy

= −1

2lnβ + terms independent of β

thus the average energy is

ε̄i = − ∂∂β

−1

2ln β

=

1

2β =

τ

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 V N ,

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 S ν( f ) but in ourcase it is convenient to write it as S V (ω) the following relationships are

S V (ω)d ω = S V ( f )d f

where

S v(ω) = S V ( f )d f

d ω

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

Q = ω0 L

R ω20 =

1

LC

we can also write Q as

Q = ω0∆ω

∆ω = R

L

Now using the equipartition we can see that there will be two square terms, thus we get the following

1

2C V 2 N =

1

2k bT V 2 N =

k BT

C

and1

2 L I 2 N =

1

2k BT I 2 N =

k BT

L

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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

I 2 N = k BT

L =

Z ∞0

S V (ω)d ω

| Z (ω)|2

this takes a voltage squared and devide by the impedence squared. We all remember that we can write

Z (ω) = R + iω L + 1

iωC

and we also rember that1

C = ω20 L

thus we find that

Z (ω) = R + iL

ω(ω− ω20)2

but

| Z (ω)|2

= L2 R2

L2 + (ω2

−ω20)

2

ω2

∆ω2

= R2

L2

this allows us to write

I 2 N =Z ∞

0

S V (ω)

L2d ω

∆ω2 + (ω2−ω20)2

ω2

this function is sharply peaked about ω0 for high Q. If we take the limit we find

S V (ω) → S V (ω0)

if we look at(ω2

−ω20)

2

ω2 =

((ω

−ω0)(ω + ω0))

2

ω2

if we let ω → ω0 we find(ω2 −ω20)2

ω2 = 4(ω −ω0)2

given this we get the following result

k BT

L=

S V (ω0)

4 L2

Z ∞0

d ω

(ω −ω0)2 + R2/4 L2 x = ω− ω0 = d ω a = R/2 L

as you can see this now looks more tractable, thus we can now write this as

S V (ω0)4 L2

Z ∞

0

dx

x2 + a2

where we all know that Z ∞0

dx

x2 + a2 =

π

a

thus we findk BT

L =

S V (ω0)π

4 L2( R/2 L) =

S V (ω0)

2 LR π

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thus we find that the spectral

S V (ω0) = 2

πk BT R

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

S V (ω) = 2πk BT R S V ( f ) = 4k bT R

this is the famous results called Nyquist noise or Johnson noise. The current noise is given by

S I ( f ) = 4k BT

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 wemust 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 ≥ k BT . As a final remark because of the linear dependence with temperature thanthe Nyquist noise is used as an absolute thermometer.

3.14 Summary

The Boltzman factor

p(εl ) = e−εl/τ

Z is the probability using the Boltzman factor. We then introduced the partition function

Z = ∑s

e−εs/τ

we then intruduced the concept of pressure, the pressure is given by

p = −

∂U

∂V

σ

= τ

∂σ

∂V

U

where we have the identity given by

dU = τd σ − 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

σ = −

dF

d τ

V

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and

p = −

dF

dV

τ

the Maxwell relations are given by ∂σ

∂V

τ

=

d p

d τ

V

where the free energy isF = −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

Z 1 = nQV nQ =

M τ

2π 2

3/2where the energy of one particle is given by

U = 3

2 τ

we then move on to the Equipartition theorem

U = τ

2

per quadratic term. Finally we talked about the Nyquist noise

S V ( f ) = 4k BT R S I ( f ) = 4k BT

R


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 therotating frame of reference is mω2r , 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 theprobability that the system is in quatum state 2 at energy ε2 is just the ratio of the two multiplicities

P(ε1)

P(ε2) =

g R(U 0 −ε1)g R(U 0 −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 g R g R = eσ

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thus Equation 1 now becomes

P(ε1)

P(ε2) =

e(σ R(U 0−ε1))

e(σ R(U 0−ε2)) (3.2)

we can define

∆σ R = σ R(U 0 − ε1) −σ R(U 0 − ε2) (3.3)this allows us to write Equation 2 as

P(ε1)

P(ε2) = e∆σ R

since we know that U 0 ≫ ε allows us to Taylor expand the entropy (Equation 3). The Taylor expansion isdefined as

f ( x0 + a) = f ( x0) + a

d f

dx

x= x0

+ 1

2!a2

d 2 f

dx2

x= x0

+ ...

thus we find

σ R(U 0 − ε1) = σ R(U 0) −ε1(∂σ R/∂U )V , N + ...

= σ R(U

0) −ε1/τσ R(U 0 − ε2) = σ R(U 0) −ε2(∂σ R/∂U )V , N + ...= σ R(U 0) −ε2/τ

thus we find

∆σ R = −(ε1 − ε2)/τand we finally find that

P(ε1)

P(ε2) = e−(ε1−ε2)/τ

thus the ratio of the number of particles in a particular quantu state 1 to the number in a particular state 2

is given by the ratios of their probabilities

N 1

N 2=

P(ε1)

P(ε2) = e−(ε1−ε2)/τ

we know that the density is defined as

ρ = m

V N

thus we know that

ρ1 = m

V N 1 ρ2 =

m

V N 2

and the relative density is given by

ρ̄ = ρ1ρ2

= N 1

N 2= e−(ε1−ε2)/τ

which is simply the Boltzman factor. Since we also know that the centrifugal force is defined as

F c = mω2r

and the energy is given by

ε = −Z

F · dr = −12

ω2r 2

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thus we know that

ρ̄ = ρ(r )

ρ(0) = eε1/τ

thus we find the relative density to be given as

ρ̄ = e1

2τ mω2r 2

a plot is given as

Problem # 2Show 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 nucleusis 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 = −τ ln Z F = U −στ

putting these two expressio equal to each other yields

−τ ln Z = U − στ

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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 = εs = ε̄

where

ε = ∑l

εl p(l) = 1

Z ∑

l

εle−εl/τ

where we often use β = 1/τ, using this we get

ε = ∑l

εl p(l) = 1

Z ∑

l

εl e−εl β

and the trick we can use∂

∂β

e−εl β =−

εl e−εl β

and now we can write

ε = 1 Z

−∂ Z

∂β

= −∂ ln Z

∂β

to put this back in terms of τ we just simply need to use the chain rule

∂ Z

∂τ =

∂ Z

∂β

∂β

∂τ =

∂ Z

∂β

− 1

τ2

thus

ε

= 1

Z τ2

∂ Z

∂τ = τ2

∂ ln Z

∂τThus the energy is given by

U = ε = 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 = ∑s

e−εs/τ

and we know the ms = ±1 εs = ε and for ms = 0 εs = 0. Thus we find thet the partition function for thissystem is given by

Z N = (1 + 2e−ε/τ) N

and the entropy of the solid is given by

σ = N ln Z +U

τ

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using Equation 4 and the partition function we find the entropy to be given as

σ = N ln(1 + 2e−ε/τ) + τ∂ ln(1 + 2e−ε/τ)

∂τ

now we just need to find∂ ln( Z N )

∂τ =

N

Z

∂ Z

∂τand we find

∂

∂τ(1 + 2e−ε/τ) = 2

∂

∂τ(e−ε/τ)

letting

u = −ετ

∂u

∂τ =

ε

τ2

thus we find

τ∂ ln(1 + 2e−ε/τ)

∂τ =

2 N ε

τ

e−ε/τ

1 + 2e−ε/τ

thus we find the total entropy to be given as

σ = N ln(1 + 2e−ε/τ) + 2 N ε

τ

e−ε/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 → ∞. Showthat your expression in (b) reduces to these values.

Since we know that the probability goes as

p = 1

g

thus the multiplicity for 1 particle goes as

g = 1

p

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and the multiplicity for N particles is given by

g( N ) =

1

p

N

and as τ → ∞ we know all the states are equally probable thus

g( N ) = 3 N p = 13

and thus at this limit the entropy is

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 theprobability is

p = 1 g = 1 N

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

limτ→0

ln(1 + 2e−ε/τ) + 2ε

τ

e−ε/τ

1 + 2e−ε/τ ≈ 0

this makes sense because we know if the temperature goes to 0 then the entropy goes 0.

In the limit that τ → ∞ we find

limτ→∞ ln(1 + 2e−ε/τ

) +

2ε

τ

e−ε/τ

1 + 2e−ε/τ ≈ ln 3which also makes sense, as the temperature goes to infinity the entropy converges to a constant number.

d) Calculate the heat capacity C V of the crystal, and find its temperature dependence for high τ. Make asketch of C V vs τ.

We know that the heat capacity is defined as

C V =

∂U

∂τ

V

at constant volume. We know that the energy is given as

U = τ2∂ ln Z

∂τ = 2 N ε

e−ε/τ

1 + 2e−ε/τ

we need to find

∂U

∂τ = 2 N ε

∂

∂τ

e−ε/τ

1 + 2e−ε/τ

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Using the quotient rule∂

∂ x

f

g

=

g f ′ − f ′gg2

if we say

f = e−ε/τ f ′ = ε

τ2e−ε/τ

andg = 1 + 2e−ε/τ g′ =

2ε

τ2e−ε/τ

thus

∂U

∂τ = 2 N ε

(1 + 2e−ε/τ)(ε/τ2)e−ε/τ −e−ε/τ(2ε/τ2)e−ε/τ

(2e−ε/τ + 1)2

this can be simplified by factoring out a term

∂U

∂τ =

2 N ε2

τ2 e−ε/τ

1 + 2e−ε/τ −2e−ε/τ

(2e−ε/τ + 1)2

=

2 N ε2

τ2e−ε/τ

(1 + 2e−ε/τ)2

thus the heat capacity for this system is given by

C V = 2 N ε2

τ2e−ε/τ

(1 + 2e−ε/τ)2

in order to sketch this function we can write it in a more suggestive form and use IDL to plot it, if we let

x = ε

τ

we can write the heat capacity as

C V = 2 Nx2 e

− x

(1 + 2e− x)2

and the plot is given as

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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 +U

τ

but we know the energy is defined as

U = τ2∂ ln Z

∂τ

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thus

σ = ln Z + τ∂ ln Z

∂τ

where

τ∂ ln Z

∂τ =

ε

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

m(τ) = 1 Z

∑l

ml e−βεl

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what are the states? there are only two states, 1 toward the field and one opposing the field

m↑ = +m ε↑ = −mB spin upm↓ = −m 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

so the expectation value is now

m(τ) = meβmB −me−βmB

eβmB + e−βmB = m

sinh(βmB)

cos(βmB) = m tanh(βmB)

thus

m(τ) = m tanh(βmB)but we are looking for the moment for all N spins

M (τ) = N m(τ) = Nm tanh

Bm

τ

but we know that

M (τ) = N m(τ)

V = nm tanh

Bm

τ

n =

N

V

thus

M (τ) = nm tanh

Bm

τ

we can also find the suceptibility to be

χ = d M

dB=

nm2

τ sech2

mB

τ

thus

χ = nm2

τ sech2

mB

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

F = −τ ln(eβmB + e−βmB) = −τ ln2cosh

mB

τ

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and to put it in terms of x ≡ M / Nm we simply do

x = M

Nm = tanh

mB

τ

thusmB

τ = tanh−1( x)

we also know that

cosh2( x) − sinh2( x) = 1 cosh( x) = 1 1 − tanh2( x)

and finally we find that

F = −τ ln( 2√ 1 − x2 )

c) Show that the susceptibility is χ = nm2/τ in the limit mB ≪ τ.

we know that for large x we find

lim x→∞ tanh( x) ≈ 1

In the limit that mB/τ �

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