The Hagedorn Temperature in String Theory

Henry ScharfUniversity of Arizona

Honors ThesisMay 3, 2007

What causes a Hagedorn temperature?

Normally a partition function goes as where p(E) is the number of different states with energy E, called the degeneracy of states.

What would happen if p(E) grew exponentially in E, and overcame the Boltzman suppression?

There is some critical value of , and hence temperature such that above it, the partition function diverges with increasing energy. We name this value H.

Z = PE p(E )e¡ ¯ E

E

EeEpZ )(

E

EEeZ )(

Probability Ratios

The probability that a system is in a given state is

Therefore the probability that a system is in any state with energy E is

It is enlightening to look at a ratio of these probabilities for arbitrary energies E1, and E2.

We can see that in the limit goes to , this ratio goes to 1. So, as we approach H, the average energy diverges.

Thus, we can never get above TH = 1/Hk.

ZeP

E

ZeEpPEP

E

EE

)()(

))((

1

2

1

2

12

)()(

)()(

EEe

eEpEp

EPEP

Quick review of thermodynamics

The entropy of a system is defined in terms of the number of states the system can be in.

The Helmholtz free energy is defined

F can be found from the partition function Z.

We can also relate S to F in the following way.

TSEF

)(ln EkS

VTFS

ZkTF

ln

Strings and the Worldsheet

To describe the path of a point particle, we assign a vector X(t) which tells us the exact location of the point for a given t. Thus X(t) describes a line through spacetime.

A string requires more information. We need to know the position of every point along the string. Thus, X will depend on two variables. These are usually called and , and they parameterize a two-dimensional object called the worldsheet.

The Nambu-Goto action

When we think of the action of a free point particle, we can think of minimizing the length of it’s worldline. This makes sense, a free particle moves in a straight line.

When we think of the action of a string, we must minimize not the length of the line, but the surface area of the worldsheet. So

f

XXXXddcTS

0

1

0

2220 )()()(

Mode expansion

The Euler-Lagrange equation for the string action ultimate yields the wave equation.

The solutions to this equation are of the form

They can be written in terms of the light-cone metric, which re-defines two coordinates.

Moreover, the n- coefficients can be

written in terms of the nI

coefficients. This means that the total information about the string is contained in the transverse oscillations.

02

2

2

2

XX

)cos(12),(

)cos(12),(

),(

)cos(12),(

00

00

00

nn

ixX

nn

ixX

pX

nn

ixX

n

In

II

nn

nn

),,(2

,2

132

1010

DI XXXX

XXXXXX

String as a SHO

Strings act like quantum simple harmonic oscillators, so we can describe them with creation/annihilation operators.

Any string state can be written in terms of creation operators acting on the vacuum, |i.

The subscripts tell us which mode they excite, and the superscripts tell us in which of the transverse directions they are acting. The exponents are called occupation numbers.

The energy of a string will depend on the number

to the extent that strings with the same N will have the same energy. So, when we consider the number of different strings with energy E, we are really considering the number of strings with constant N.

)()(2

)(1 )()()( )()(

2)(

1

Kk

JI nKk

nJnIaaa

1

1

2

)(

D

q

qnN

Partitions on N

For now, we will simplify things by considering the case of 1 transverse direction, or D-2 = 1.

Ultimately, we want to know the number of ways to choose occupation numbers nl such that the total sum

remains constant.

This amounts to asking the number of ways to add up positive integers to get N.

This number p(N) is known as the number of partitions on N

1

nN

Table of p(N)

An estimate for p(N) when N is large

Consider a one-dimensional quantized vibrating string. It will have some discrete set of vibrational modes, which can be described by occupation numbers.

If we define N as shown, such a system will have energy E.

Our strategy for estimating p(N) will be to examine this system, derive an expression for ln[p(N)] using our knowledge of thermodynamics and statistics.

1

nN

knk

nn aaa )()()( 2121

yyy

NE 0

Partition function

Looking at partition function, we sum over all possible states.

This is the same as summing over all possible values for the nl independently.

We can split this sum up over the n

l.

We can see that each of the terms in Z is geometric.

,,

)32(

21

3210

nn

nnneZ

E

EeZ

2

20

1

10 2

n

n

n

n eeZ

n

n

Z )][exp(1 0

0

An approximation

Now we have: Remember the following

equation concerning the Helmholtz free energy:

Plugging in Z, we get an expression for F as a sum.

We can write this as an integral since the terms don’t change quickly.

We can then Taylor-expand the natural log to get

1

0 )]exp(1ln[

kTF

1

10 ])exp(1[

Z

ZkTF ln

)1(00

1

0 )]exp(1ln[ dkTF

1 1

0 )exp(n n

ndkTF

The entropy as a function of N

The sum in F can be pulled outside of the integral, and the integral evaluated to give

The sum over n-2 is well known. Now recalling another equation from

thermodynamics, we can find the entropy, S.

Using the definition of the Helmholtz free energy, we can write S as a function of E.

Finally, remembering the relationship between E and N, we can write S as a function of N.

1

20

2 1)(n n

kTF

6)( 2

0

2

kTF

0

2

3)(

TkTFS

V

062

EkS

62 NkS

Final estimate for ln[pD-2(N)]

Now, we only need to look at the definition of entropy to find our estimate.

The number of choices we have for our occupation numbers will actually go up by a factor of D-2.

The new partition function is ZD-2 and it can be shown that ZD-2= ZD-2.

Since the entropy depends on ln(ZD-2), it’s not surprising that SD-2= (D-2)S.

Ultimately, this means that our estimate for p(N) picks up a factor of (D-2)1/2

62)(ln

)(ln)(ln)(

NNp

NpkEkES

6)2(2)(ln 2NDNpD

The string partition function

Unlike the previous case, the energy of a string depends on the square root of N, henceforth called N?.

So now we can write down the partition function for an open string.

NE

NE 112

N

E

N

NNDZ

eNpZ N

6)2(2exp

)(

Hagedorn temperature

It’s easy to see that the partition function diverges for values of greater than some critical point.

This H defines a corresponding temperature, the Hagedorn temperature.

απkT

kT

NND

H

H

H

41

16

)226(2

6)2(2

Another explanation

The Hagedorn temperature is not necessarily a maximum temperature.

The details of what actually happens near TH turn out to depend strongly on what happens to the potential energy of the string U(T) as T goes to TH.

As it turns out, what happens at TH is a phase change. Exactly how we can interpret this is the subject of much speculation.

Acknowledgements

The structure of the derivation of the Hagedorn temperature in Chapter 4, along with much of the preparation in preceding chapters came directly from Zweibach. I thank him and his well-prepared text without which I would have been lost. It is a unique tool which introduces to undergraduates the beginnings of string theory in a rigorous way.

Thanks to Mike Lennek, who offered his assistance and expertise in the study of the Hagedorn temperature in strings.

Thanks also to my adviser, Dr. Keith R. Dienes, who directed my study over the past year. Beginning with my first semester at the University of Arizona four years ago, he has provided me with questions, answers, and an ever deepening curiosity about the physical universe every time I sit down in his office.