Download Entropy, Strings, and Partitions of Integers

Entropy, Strings, and Partitions of Integers
Jonathan Tannenhauser
ANOMALY
Wellesley College
November 4, 2010
1
1. Preliminaries
A partition of a positive integer N is a way of writing N as a sum
of positive integers.
1. Preliminaries
A partition of a positive integer N is a way of writing N as a sum
of positive integers.
How many partitions does the positive integer N have?
Call this number p(N).
1. Preliminaries
A partition of a positive integer N is a way of writing N as a sum
of positive integers.
How many partitions does the positive integer N have?
Call this number p(N).
Example. p(4) = 5:
I
4
I
3+1
I
2+2
I
2+1+1
I
1+1+1+1
There’s a formula for p(N):
…
∞
1 X
√
d
p(N) = √
Ak (N) k
dN
π 2 k=1
where
X
Ak (N) =
sinh
q €
π
2
k
È
3
N−
N−
1
24
1
24
Š 
,
e πi(s(m,k)−2Nm/k)
0≤m<k , (m,k)=1
and
s(m, k) =
k−1
X
j=1
j
k
jm
jm
1
−
−
k
k
2
.
(Rademacher 1937)
3
In 1918, Hardy and Ramanujan proved an asymptotic formula for
p(N):
For large N,
1
√ exp 2π
p(N) ∼
4N 3
Ê !
N
6
4
In 1918, Hardy and Ramanujan proved an asymptotic formula for
p(N):
For large N,
1
√ exp 2π
p(N) ∼
4N 3
Ê !
N
6
Our Goal: To prove (the red part of) this formula.
In 1918, Hardy and Ramanujan proved an asymptotic formula for
p(N):
For large N,
1
√ exp 2π
p(N) ∼
4N 3
Ê !
N
6
Our Goal: To prove (the red part of) this formula.
Ê
ln p(N) ∼ 2π
N
.
6
Proposition. (Euler)
∞
X
N=0
p(N)x N =
∞
Y
1
.
1 − x`
`=1
5
Proposition. (Euler)
∞
X
p(N)x N =
N=0
∞
Y
1
.
1 − x`
`=1
Proof. Expand the right-hand side:
(1 + x 1 + (x 1 )2 + · · · ) (1 + x 2 + (x 2 )2 + · · · ) (1 + x 3 + (x 3 )2 · · · ) · · · .
`=1
`=2
`=3
5
Proposition. (Euler)
∞
X
p(N)x N =
N=0
∞
Y
1
.
1 − x`
`=1
Proof. Expand the right-hand side:
(1 + x 1 + (x 1 )2 + · · · ) (1 + x 2 + (x 2 )2 + · · · ) (1 + x 3 + (x 3 )2 · · · ) · · · .
`=1
`=2
`=3
Each partition of N gives one way of realizing x N in the product.
5
Proposition. (Euler)
∞
X
p(N)x N =
N=0
∞
Y
1
.
1 − x`
`=1
Proof. Expand the right-hand side:
(1 + x 1 + (x 1 )2 + · · · ) (1 + x 2 + (x 2 )2 + · · · ) (1 + x 3 + (x 3 )2 · · · ) · · · .
`=1
`=2
`=3
Each partition of N gives one way of realizing x N in the product.
For example, the partition 7 = 1 + 1 + 2 + 3 is realized by the blue
terms.
5
Proposition. (Euler)
∞
X
p(N)x N =
N=0
∞
Y
1
.
1 − x`
`=1
Proof. Expand the right-hand side:
(1 + x 1 + (x 1 )2 + · · · ) (1 + x 2 + (x 2 )2 + · · · ) (1 + x 3 + (x 3 )2 · · · ) · · · .
`=1
`=2
`=3
Each partition of N gives one way of realizing x N in the product.
For example, the partition 7 = 1 + 1 + 2 + 3 is realized by the blue
terms.
Therefore, the coefficient of x N in the product is the number of
partitions of N.
5
2. The Quantum String
6
2. The Quantum String
A harmonic oscillator is a system that vibrates back and forth at
some frequency ω, such as a mass on a spring.
6
2. The Quantum String
A harmonic oscillator is a system that vibrates back and forth at
some frequency ω, such as a mass on a spring.
In classical mechanics, the energy of an oscillator can be any
non-negative number.
6
2. The Quantum String
A harmonic oscillator is a system that vibrates back and forth at
some frequency ω, such as a mass on a spring.
In classical mechanics, the energy of an oscillator can be any
non-negative number.
In quantum mechanics, the only permitted energies are of the form
En = n~ω ,
where n is a non-negative integer and ~ = 1.05 × 10−34 J·s is
Planck’s constant.
6
2. The Quantum String
A harmonic oscillator is a system that vibrates back and forth at
some frequency ω, such as a mass on a spring.
In classical mechanics, the energy of an oscillator can be any
non-negative number.
In quantum mechanics, the only permitted energies are of the form
En = n~ω ,
where n is a non-negative integer and ~ = 1.05 × 10−34 J·s is
Planck’s constant.
Terminology: An oscillator with energy En “is in state n” or
“has n excitations at frequency ω.”
6
A classical string has some fundamental frequency ω, set by the
requirement that the ends of the string are fixed.
7
A classical string has some fundamental frequency ω, set by the
requirement that the ends of the string are fixed.
The string’s allowed frequencies of vibration are positive integer
multiples of this fundamental frequency:
ω, 2ω, 3ω, . . . .
7
So we can think of a string as being made up of an infinite
collection of harmonic oscillators, with frequencies ω, 2ω, 3ω, . . . .
So we can think of a string as being made up of an infinite
collection of harmonic oscillators, with frequencies ω, 2ω, 3ω, . . . .
Any pattern of vibration can be decomposed as a linear
combination of these basic modes of vibration,
y (x) = a1 sin ωx + a2 sin 2ωx + a3 sin 3ωx + · · · .
A quantum string is a collection of oscillators with frequencies `ω,
where ` is a positive integer.
A quantum string is a collection of oscillators with frequencies `ω,
where ` is a positive integer.
Each oscillator has a discrete series of allowed energies
E`,n` = n` ~(`ω) .
(n` = 0, 1, 2, . . . .)
A quantum string is a collection of oscillators with frequencies `ω,
where ` is a positive integer.
Each oscillator has a discrete series of allowed energies
E`,n` = n` ~(`ω) .
(n` = 0, 1, 2, . . . .)
Therefore, the state of a quantum string is specified by a sequence
ψ of non-negative integers,
ψ = (n1 , n2 , n3 , . . . ) ,
where n` is the state number of the oscillator with frequency `ω.
A quantum string is a collection of oscillators with frequencies `ω,
where ` is a positive integer.
Each oscillator has a discrete series of allowed energies
E`,n` = n` ~(`ω) .
(n` = 0, 1, 2, . . . .)
Therefore, the state of a quantum string is specified by a sequence
ψ of non-negative integers,
ψ = (n1 , n2 , n3 , . . . ) ,
where n` is the state number of the oscillator with frequency `ω.
The total energy of a string in state ψ is the sum of the energies in
each of the individual oscillators,
E = n1 ~ω + n2 ~(2ω) + n3 ~(3ω) + · · · ≡ N~ω ,
where
N = n1 + 2n2 + 3n3 + · · · .
There is a correspondence between states ψ of the string and
partitions of N = n1 + 2n2 + 3n3 + · · · :
The state ψ = (n1 , n2 , n3 , . . . ) corresponds to the partition
z
n1 summands
}|
{
z
n2 summands
}|
{
z
n3 summands
}|
{
N = 1 + 1 + ··· + 1+2 + 2 + ··· + 2+3 + 3 + ··· + 3+··· .
10
There is a correspondence between states ψ of the string and
partitions of N = n1 + 2n2 + 3n3 + · · · :
The state ψ = (n1 , n2 , n3 , . . . ) corresponds to the partition
z
n1 summands
}|
{
z
n2 summands
}|
{
z
n3 summands
}|
{
N = 1 + 1 + ··· + 1+2 + 2 + ··· + 2+3 + 3 + ··· + 3+··· .
So if we can count how many string states there are with energy
N~ω, we’ll have solved our partition problem!
10
3. Statistical Mechanics
11
3. Statistical Mechanics
In classical physics, if we know the position and velocity of every
particle in a system at a given time, then we can calculate the
course of that system forever after.
11
3. Statistical Mechanics
In classical physics, if we know the position and velocity of every
particle in a system at a given time, then we can calculate the
course of that system forever after.
In practice, macroscopic systems contain too many particles to
make such a calculation feasible.
11
3. Statistical Mechanics
In classical physics, if we know the position and velocity of every
particle in a system at a given time, then we can calculate the
course of that system forever after.
In practice, macroscopic systems contain too many particles to
make such a calculation feasible.
But what we really care about is not the microscopic behavior of
each of the constituent particles, but the macroscopic features of
the system—its average energy, its temperature, its pressure.
11
3. Statistical Mechanics
In classical physics, if we know the position and velocity of every
particle in a system at a given time, then we can calculate the
course of that system forever after.
In practice, macroscopic systems contain too many particles to
make such a calculation feasible.
But what we really care about is not the microscopic behavior of
each of the constituent particles, but the macroscopic features of
the system—its average energy, its temperature, its pressure.
These features depend only on the aggregate properties of the
particles, so we can use statistical methods to study them.
11
A. Entropy
12
A. Entropy
Consider an isolated physical system in a particular macroscopic
state.
12
A. Entropy
Consider an isolated physical system in a particular macroscopic
state.
There are many microstates—call their number W —that give rise
to this macrostate.
A. Entropy
Consider an isolated physical system in a particular macroscopic
state.
There are many microstates—call their number W —that give rise
to this macrostate.
The entropy of the system is defined as
S = k ln W ,
where k = 1.38 × 10−23 J/K is Boltzmann’s constant.
12
13
In writing S = k ln W , we have assumed each microstate is equally
probable.
14
In writing S = k ln W , we have assumed each microstate is equally
probable.
In many systems, the microstates that give rise to the same
macrostate are not equally probable.
14
In writing S = k ln W , we have assumed each microstate is equally
probable.
In many systems, the microstates that give rise to the same
macrostate are not equally probable.
If the probability of microstate ψ is pψ , then we define the
system’s entropy to be
S = −k
X
pψ ln pψ .
ψ
14
In writing S = k ln W , we have assumed each microstate is equally
probable.
In many systems, the microstates that give rise to the same
macrostate are not equally probable.
If the probability of microstate ψ is pψ , then we define the
system’s entropy to be
S = −k
X
pψ ln pψ .
ψ
Note. For a system with W equally probable microstates,
pψ = 1/W for all ψ, so that
S = −k
W
X
1
ψ=1
W
ln
1‹
W
= −kW
1‹
W
(− ln W ) = k ln W .
14
B. The Partition Function
B. The Partition Function
Consider a non-isolated system, at fixed temperature T , that is
free to exchange energy with its surroundings, so that its energy is
not fixed.
B. The Partition Function
Consider a non-isolated system, at fixed temperature T , that is
free to exchange energy with its surroundings, so that its energy is
not fixed.
We can achieve these conditions by keeping our system in contact
with a heat bath at temperature T .
B. The Partition Function
Consider a non-isolated system, at fixed temperature T , that is
free to exchange energy with its surroundings, so that its energy is
not fixed.
We can achieve these conditions by keeping our system in contact
with a heat bath at temperature T .
What is the probability pψ of finding our system in microstate ψ
with energy Eψ ?
Imagine an ensemble consisting of many copies of our system in
contact with the heat bath.
Imagine an ensemble consisting of many copies of our system in
contact with the heat bath.
Let
aψ1
aψ2
aψ3
be the number of copies in state
ψ1
ψ2
ψ3
with energy
Eψ1 ,
Eψ2 ,
Eψ3 .
17
Let
Then
aψ1
aψ2
aψ3
be the number of copies in state
ψ1
ψ2
ψ3
with energy
Eψ1 ,
Eψ2 ,
Eψ3 .
aψ2
= f (Eψ1 , Eψ2 ) = f (Eψ2 − Eψ1 ).
aψ1
17
Let
aψ1
aψ2
aψ3
be the number of copies in state
ψ1
ψ2
ψ3
Then
aψ2
= f (Eψ1 , Eψ2 ) = f (Eψ2 − Eψ1 ).
aψ1
Similarly,
aψ3
= f (Eψ3 − Eψ2 )
aψ2
and
aψ3
= f (Eψ3 − Eψ1 ).
aψ1
with energy
Eψ1 ,
Eψ2 ,
Eψ3 .
Let
aψ1
aψ2
aψ3
be the number of copies in state
ψ1
ψ2
ψ3
Then
aψ2
= f (Eψ1 , Eψ2 ) = f (Eψ2 − Eψ1 ).
aψ1
Similarly,
aψ3
= f (Eψ3 − Eψ2 )
aψ2
and
aψ3
= f (Eψ3 − Eψ1 ).
aψ1
Since
aψ3
aψ aψ
= 2 · 3,
aψ1
aψ1 aψ2
with energy
f (Eψ3 − Eψ1 ) = f (Eψ2 − Eψ1 )f (Eψ3 − Eψ2 ) .
Eψ1 ,
Eψ2 ,
Eψ3 .
The only function for which
f (Eψ3 − Eψ1 ) = f (Eψ2 − Eψ1 )f (Eψ3 − Eψ2 )
holds for all Eψi is an exponential function.
18
The only function for which
f (Eψ3 − Eψ1 ) = f (Eψ2 − Eψ1 )f (Eψ3 − Eψ2 )
holds for all Eψi is an exponential function.
Therefore, for each state ψ,
f (Eψ ) ∝ exp(−βEψ ) ,
for some constant β, and likewise
aψ ∝ exp(−βEψ ) .
18
The only function for which
f (Eψ3 − Eψ1 ) = f (Eψ2 − Eψ1 )f (Eψ3 − Eψ2 )
holds for all Eψi is an exponential function.
Therefore, for each state ψ,
f (Eψ ) ∝ exp(−βEψ ) ,
for some constant β, and likewise
aψ ∝ exp(−βEψ ) .
But aψ ∝ pψ , the probability of finding the system in state ψ.
18
The only function for which
f (Eψ3 − Eψ1 ) = f (Eψ2 − Eψ1 )f (Eψ3 − Eψ2 )
holds for all Eψi is an exponential function.
Therefore, for each state ψ,
f (Eψ ) ∝ exp(−βEψ ) ,
for some constant β, and likewise
aψ ∝ exp(−βEψ ) .
But aψ ∝ pψ , the probability of finding the system in state ψ.
Therefore, pψ ∝ exp(−βEψ ) .
18
Fact. β = 1/kT : that is,
pψ ∝ exp −
Eψ
kT
.
Fact. β = 1/kT : that is,
pψ ∝ exp −
Eψ
kT
.
Intuition. kT is the characteristic energy scale of the system.
States with energy . kT are probable; states with energy & kT
are improbable.
Fact. β = 1/kT : that is,
pψ ∝ exp −
Eψ
kT
.
Intuition. kT is the characteristic energy scale of the system.
States with energy . kT are probable; states with energy & kT
are improbable.
If we call the constant of proportionality 1/Z , then
Fact. β = 1/kT : that is,
pψ ∝ exp −
Eψ
kT
.
Intuition. kT is the characteristic energy scale of the system.
States with energy . kT are probable; states with energy & kT
are improbable.
If we call the constant of proportionality 1/Z , then
Eψ
1
pψ = exp −
Z
kT
.
Fact. β = 1/kT : that is,
pψ ∝ exp −
Eψ
kT
.
Intuition. kT is the characteristic energy scale of the system.
States with energy . kT are probable; states with energy & kT
are improbable.
If we call the constant of proportionality 1/Z , then
Eψ
1
pψ = exp −
Z
kT
.
This probability distribution is called the Boltzmann distribution.
In the expression
Eψ
1
pψ = exp −
Z
kT
,
we can determine Z by noting that our system must be in some
microstate, so
X
ψ
pψ =
Eψ
1X
exp −
= 1.
Z ψ
kT
In the expression
Eψ
1
pψ = exp −
Z
kT
,
we can determine Z by noting that our system must be in some
microstate, so
X
ψ
pψ =
Eψ
1X
exp −
= 1.
Z ψ
kT
So
Z=
X
ψ
Eψ
exp −
kT
.
In the expression
Eψ
1
pψ = exp −
Z
kT
,
we can determine Z by noting that our system must be in some
microstate, so
X
ψ
pψ =
Eψ
1X
exp −
= 1.
Z ψ
kT
So
Z=
X
ψ
Eψ
exp −
kT
.
The factor Z is called the partition function. It enables us to
compute any macroscopic property of our system.
C. Average Energy, Entropy, and Free Energy
21
C. Average Energy, Entropy, and Free Energy
We can now compute various physically important macroscopic
quantities, starting from
pψ =
Eψ
1
exp −
Z
kT
(probability of microstate ψ)
and
Z=
X
ψ
Eψ
exp −
kT
(the partition function)
21
C. Average Energy, Entropy, and Free Energy
We can now compute various physically important macroscopic
quantities, starting from
pψ =
Eψ
1
exp −
Z
kT
(probability of microstate ψ)
and
Z=
X
ψ
Eψ
exp −
kT
(the partition function)
The system’s average energy hE i is the expected value of Eψ ,
hE i =
X
ψ
Eψ
1X
pψ Eψ =
exp −
Eψ .
Z ψ
kT
Let us compute the entropy
S = −k
X
ψ
pψ ln pψ .
Let us compute the entropy
S = −k
X
pψ ln pψ .
ψ
Since
we find
Eψ
1
pψ = exp −
Z
kT
,
Let us compute the entropy
S = −k
X
pψ ln pψ .
ψ
Since
Eψ
1
pψ = exp −
Z
kT
,
we find
S = −k
X
ψ
pψ
Eψ
−
− ln Z
kT
=
hE i
+ k ln Z .
T
Let us compute the entropy
S = −k
X
pψ ln pψ .
ψ
Since
Eψ
1
pψ = exp −
Z
kT
,
we find
S = −k
X
ψ
pψ
Eψ
−
− ln Z
kT
=
hE i
+ k ln Z .
T
Rearranging, we get
hE i − TS = −kT ln Z ≡ F ,
which defines the free energy F .
Let us compute the entropy
S = −k
X
pψ ln pψ .
ψ
Since
Eψ
1
pψ = exp −
Z
kT
,
we find
S = −k
X
ψ
pψ
Eψ
−
− ln Z
kT
=
hE i
+ k ln Z .
T
Rearranging, we get
hE i − TS = −kT ln Z ≡ F ,
which defines the free energy F .
Fact. S = −
dF
dT
22
4. The Main Calculation
4. The Main Calculation
A quantum string in state ψ = (n1 , n2 , n3 , . . . ) with
n1 excitations at frequency ω,
n2 excitations at frequency 2ω,
n3 excitations at frequency 3ω, etc.,
has energy
Eψ = ~ωN ,
where
N = n1 + 2n2 + 3n3 + · · · =
∞
X
`=1
`n` .
4. The Main Calculation
A quantum string in state ψ = (n1 , n2 , n3 , . . . ) with
n1 excitations at frequency ω,
n2 excitations at frequency 2ω,
n3 excitations at frequency 3ω, etc.,
has energy
Eψ = ~ωN ,
where
N = n1 + 2n2 + 3n3 + · · · =
∞
X
`n` .
`=1
This string state corresponds to the partition
z
n1 summands
}|
{
z
n2 summands
}|
{
z
n3 summands
}|
{
N = 1 + 1 + ··· + 1+2 + 2 + ··· + 2+3 + 3 + ··· + 3+··· .
23
Therefore, counting the number of partitions of N is equivalent to
counting the number of string states with energy N~ω.
24
Therefore, counting the number of partitions of N is equivalent to
counting the number of string states with energy N~ω.
But (k times the logarithm of) the number of the string states is
the entropy.
Therefore, counting the number of partitions of N is equivalent to
counting the number of string states with energy N~ω.
But (k times the logarithm of) the number of the string states is
the entropy.
Strategy: Use statistical mechanics to calculate the entropy of the
string, and thereby to count the number of partitions of N.
24
Everything begins with the partition function
Z
=
X
ψ
Eψ
exp −
kT
25
Everything begins with the partition function
Z
=
X
ψ
=
Eψ
exp −
kT
∞ X
∞
X
n1 =0 n2
~ω
· · · exp −
(n1 + 2n2 + · · · )
kT
=0
25
Everything begins with the partition function
Z
=
X
ψ
=
Eψ
exp −
kT
∞ X
∞
X
n1 =0 n2
2
~ω
· · · exp −
(n1 + 2n2 + · · · )
kT
=0
32
3
∞
X
~ω
~ω
5
4
4
· n1
· 2n2 5 · · ·
=
exp −
exp −
kT
kT
n =0
n =0
∞
X
2
3
∞
∞
Y
X
~ω
4
=
exp −
· `n` 5
1
`=1
2
n` =0
kT
25
Everything begins with the partition function
Z
=
X
ψ
=
Eψ
exp −
kT
∞ X
∞
X
n1 =0 n2
2
~ω
· · · exp −
(n1 + 2n2 + · · · )
kT
=0
32
3
∞
X
~ω
~ω
5
4
4
· n1
· 2n2 5 · · ·
=
exp −
exp −
kT
kT
n =0
n =0
∞
X
2
3
∞
∞
Y
X
~ω
4
=
exp −
· `n` 5
1
=
`=1 n` =0
∞ Y
`=1
2
kT
~ω
1 − exp −
·`
kT
−1
25
Everything begins with the partition function
Z
=
X
ψ
=
Eψ
exp −
kT
∞ X
∞
X
n1 =0 n2
2
~ω
· · · exp −
(n1 + 2n2 + · · · )
kT
=0
32
3
∞
X
~ω
~ω
5
4
4
· n1
· 2n2 5 · · ·
=
exp −
exp −
kT
kT
n =0
n =0
∞
X
2
3
∞
∞
Y
X
~ω
4
=
exp −
· `n` 5
1
=
`=1 n` =0
∞ Y
`=1
2
kT
~ω
1 − exp −
·`
kT
−1
∞
X
1
Note. If we define x = e
, then Z =
=
p(N)x N
`
1
−
x
`=1
N=0
is exactly the generating function for partitions!
− ~ω
kT
∞
Y
25
The free energy is
F
= −kT ln Z
26
The free energy is
F
= −kT ln Z
= kT ln
∞ Y
`=1
~ω
1 − exp −
·`
kT
−1
26
The free energy is
F
= −kT ln Z
= kT ln
∞ Y
`=1
~ω
1 − exp −
·`
kT
∞
X
~ω`
= kT
ln 1 − exp −
kT
`=1
−1
.
26
The free energy is
F
= −kT ln Z
= kT ln
∞ Y
`=1
~ω
1 − exp −
·`
kT
∞
X
~ω`
= kT
ln 1 − exp −
kT
`=1
−1
.
We now take the limit T → ∞, or equivalently, set
~ω
1.
kT
We’ll see later that this is tantamount to assuming N 1.
26
The free energy is
F
= −kT ln Z
= kT ln
∞ Y
`=1
~ω
1 − exp −
·`
kT
∞
X
~ω`
= kT
ln 1 − exp −
kT
`=1
−1
.
We now take the limit T → ∞, or equivalently, set
~ω
1.
kT
We’ll see later that this is tantamount to assuming N 1.
”
€
In this limit, the summand ln 1 − exp − ~ω`
kT
with `.
Š—
changes slowly
26
So we may approximate the sum by an integral,
Z∞ ~ω`
F ≈ kT
ln 1 − exp −
d` .
kT
0
27
So we may approximate the sum by an integral,
Z∞ ~ω`
F ≈ kT
ln 1 − exp −
d` .
kT
0
Changing variables to q =
~ω`
kT ,
27
So we may approximate the sum by an integral,
Z∞ ~ω`
F ≈ kT
ln 1 − exp −
d` .
kT
0
Changing variables to q =
~ω`
kT ,
(kT )2
F ≈
~ω
Z
∞
ln(1 − e −q ) dq .
0
27
So we may approximate the sum by an integral,
Z∞ ~ω`
F ≈ kT
ln 1 − exp −
d` .
kT
0
Changing variables to q =
~ω`
kT ,
(kT )2
F ≈
~ω
Z
∞
ln(1 − e −q ) dq .
0
Taylor-expanding the logarithm via
‚
y2 y3
ln(1 − y ) = − y +
+
+ ···
2
3
Œ
,
27
So we may approximate the sum by an integral,
Z∞ ~ω`
F ≈ kT
ln 1 − exp −
d` .
kT
0
Changing variables to q =
~ω`
kT ,
(kT )2
F ≈
~ω
Z
∞
ln(1 − e −q ) dq .
0
Taylor-expanding the logarithm via
‚
y2 y3
ln(1 − y ) = − y +
+
+ ···
2
3
F
≈ −
(kT )2
~ω
Z
∞
0
Œ
,
(e −q + 12 e −2q + 13 e −3q + · · · ) dq
27
So we may approximate the sum by an integral,
Z∞ ~ω`
F ≈ kT
ln 1 − exp −
d` .
kT
0
Changing variables to q =
~ω`
kT ,
(kT )2
F ≈
~ω
Z
∞
ln(1 − e −q ) dq .
0
Taylor-expanding the logarithm via
‚
y2 y3
ln(1 − y ) = − y +
+
+ ···
2
3
Œ
,
Z
F
(kT )2 ∞ −q 1 −2q 1 −3q
(e + 2 e
+ 3e
+ · · · ) dq
~ω 0

‹
(kT )2
1
1
= −
1 + 2 + 2 + ···
~ω
2
3
≈ −
27
So we may approximate the sum by an integral,
Z∞ ~ω`
F ≈ kT
ln 1 − exp −
d` .
kT
0
Changing variables to q =
~ω`
kT ,
(kT )2
F ≈
~ω
Z
∞
ln(1 − e −q ) dq .
0
Taylor-expanding the logarithm via
‚
y2 y3
ln(1 − y ) = − y +
+
+ ···
2
3
F
(kT )2
~ω
(kT )2
= −
~ω
(kT )2
= −
~ω
Z
∞
≈ −
0

,
(e −q + 12 e −2q + 13 e −3q + · · · ) dq
1+
·
Œ
π2
.
6
1
1
+
+ ···
22 32
‹
27
The entropy is
d
dF
=−
S =−
dT
dT
‚
(kT )2 π 2
−
·
~ω
6
Œ
=
k 2T π2
·
.
~ω
3
28
The entropy is
d
dF
=−
S =−
dT
dT
‚
(kT )2 π 2
−
·
~ω
6
Œ
=
k 2T π2
·
.
~ω
3
The average energy is
hE i = F + TS
(kT )2 π 2 (kT )2 π 2
·
+
·
= −
~ω
6
~ω
3
2
2
π
kT
=
· ~ω
6 ~ω
The entropy is
d
dF
=−
S =−
dT
dT
‚
(kT )2 π 2
−
·
~ω
6
Œ
=
k 2T π2
·
.
~ω
3
The average energy is
hE i = F + TS
(kT )2 π 2 (kT )2 π 2
·
+
·
= −
~ω
6
~ω
3
2
2
π
kT
=
· ~ω
6 ~ω
Since E = N~ω, we can now see that the large-T limit is the same
as the large-N limit.
The entropy is
d
dF
=−
S =−
dT
dT
‚
(kT )2 π 2
−
·
~ω
6
Œ
=
k 2T π2
·
.
~ω
3
The average energy is
hE i = F + TS
(kT )2 π 2 (kT )2 π 2
·
+
·
= −
~ω
6
~ω
3
2
2
π
kT
=
· ~ω
6 ~ω
Since E = N~ω, we can now see that the large-T limit is the same
as the large-N limit.
In this limit, the string is overwhelmingly likely to be in a state
near its average energy, so we may freely interchange E and hE i.
The entropy is
d
dF
=−
S =−
dT
dT
‚
(kT )2 π 2
−
·
~ω
6
Œ
=
k 2T π2
·
.
~ω
3
The average energy is
hE i = F + TS
(kT )2 π 2 (kT )2 π 2
·
+
·
= −
~ω
6
~ω
3
2
2
π
kT
=
· ~ω
6 ~ω
Since E = N~ω, we can now see that the large-T limit is the same
as the large-N limit.
In this limit, the string is overwhelmingly likely to be in a state
near its average energy, so we may freely interchange E and hE i.
Written as a function of the energy, the entropy is
Ê
S(E ) = kπ
2E
.
3~ω
Ê
Substituting E = N~ω into S(E ) = kπ
2E
, we get
3~ω
29
Ê
Substituting E = N~ω into S(E ) = kπ
Ê
S = k · 2π
2E
, we get
3~ω
N
.
6
29
Ê
Substituting E = N~ω into S(E ) = kπ
Ê
S = k · 2π
2E
, we get
3~ω
N
.
6
On the other hand, S = k ln W , where W is the number of states
of the string;
29
Ê
Substituting E = N~ω into S(E ) = kπ
Ê
S = k · 2π
2E
, we get
3~ω
N
.
6
On the other hand, S = k ln W , where W is the number of states
of the string;
and W = p(N);
29
Ê
Substituting E = N~ω into S(E ) = kπ
Ê
S = k · 2π
2E
, we get
3~ω
N
.
6
On the other hand, S = k ln W , where W is the number of states
of the string;
and W = p(N);
Ê
so
k ln p(N) = k · 2π
N
,
6
29
Ê
Substituting E = N~ω into S(E ) = kπ
Ê
S = k · 2π
2E
, we get
3~ω
N
.
6
On the other hand, S = k ln W , where W is the number of states
of the string;
and W = p(N);
Ê
so
k ln p(N) = k · 2π
or
Ê
ln p(N) = 2π
N
,
6
N
,
6
Q.E .D.!
29