Transcript scharf - University of Arizona

The Hagedorn Temperature in
String Theory
Henry Scharf
University of Arizona
Honors Thesis
May 3, 2007
What causes a Hagedorn
temperature?


Z   p( E )e bE
Normally a partition function goes as
E
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?
Z   e(E bE )
E

There is some critical value of b, and hence
temperature such that above it, the partition
function diverges with increasing energy. We
name this value bH.
Probability Ratios





 bE
e 
The probability that a system is in a
P 
given state  is
Z
Therefore the probability that a system
p( E )e  bE
is in any state with energy E is
P( E ) 
P 
Z
E  E
It is enlightening to look at a ratio of
these probabilities for arbitrary energies
E1, and E2.
P( E2 ) p( E2 )  b

e
We can see that in the limit b goes to ,
P( E1 ) p( E1 )
this ratio goes to 1. So, as we approach
(  b )( E2  E1 )

e
bH, the average energy diverges.
Thus, we can never get above TH =
1/bHk.

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.
S  k ln ( E )
F  E  TS
F  kT ln Z
F
S 
T V
Strings and the Worldsheet


To describe the path of a point particle, we assign a vector
Xm(t) which tells us the exact location of the point for a
given t. Thus Xm(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 s and t,
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
T0
S 
c
tf
s1
t0
0
  X ) 2  ( X ) 2 ( X ) 2
d
t
d
s
(
X
 
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 redefines 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.
m
m
 X
 X

0
2
2
t
s
2
2
1
X m (t , s )  x0m  i 2    nm cos(ns )
n0 n
X  (t , s )  b p t
1
X  (t , s )  x0  i 2    n cos(ns )
n0 n
1
X I (t , s )  x0I  i 2    nI cos(ns )
n0 n
X 0  X1  X 0  X1
X 
,X 
2
2
X I  ( X 2 , X 3 , X D1 )

String as a SHO

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

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.
(K )
(J )
( I ) n ( I )
( J ) n2
( K ) nk
1
  (a1 ) (a2 ) (ak )  
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.
 D 1
The energy of a string will depend on the number N 
n ( q )

 1 q  2
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.

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

N   n


 1
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
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.
  (a1y )n1 (a2y )n2 (aky )nk  

N   n
 1
E  0 N
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 nl.
Z
We can see that each of the
terms in Z is geometric.
Z  e
 bE
E
Z
 b0 ( n1  2 n2 3n3 )
e

n1 ,n2 ,
  e b0n1   e b0 2n2 
n1
n2


Z   [exp(b0)]
n
 1 n 0
An approximation
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


Now we have:
Z  [1  exp(b0)]1
Remember the following
 1
equation concerning the
F  kT ln Z
Helmholtz free energy:

Plugging in Z, we get an
F  kT ln[1  exp( b0)]
expression for F as a sum.
 1
We can write this as an integral
b0  b0 (  1)
since the terms don’t change

quickly.
We can then Taylor-expand the F  kT d ln[1  exp(b0)]
1
natural log to get




 exp(nb0)
F  kT  d
n
n 1
1
The entropy as a function of N

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


2 
(
kT
)
1
The sum in F can be pulled outside of F  

the integral, and the integral
0 n1 n 2
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.
(kT ) 2  2
F 
0 6
F
(k ) 2 T
S 

T V
30
2k E
S
6 0
S  2k
N
6
Final estimate for ln[pD-2(N)]

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
Now, we only need to look at the
definition of entropy to find our
S ( E )  k ln ( E )
estimate.
 k ln p( N )
The number of choices we have for our
occupation numbers will actually go up
N
by a factor of D-2.
ln p( N )  2
6
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), ln p ( N )  2 ( D  2) N
D2
it’s not surprising that SD-2= (D-2)S.
6
Ultimately, this means that our estimate
for p(N) picks up a factor of (D-2)1/2
The string partition function


1
2
Unlike the previous
E  N   1

case, the energy of a
string depends on the
N
E
square root of N,

?
henceforth called N .
So now we can write
 bE
Z   p ( N  )e
down the partition
N
function for an open


(
D

2
)
N
Z   exp 2
b
string.

6
N

N

N  
  
Hagedorn temperature


It’s easy to see that the
partition function diverges
for values of b greater
than some critical point.
This bH defines a
corresponding
temperature, the
Hagedorn temperature.
( D  2) N 
2
 bH
6
N

(26  2) 
1
2

6
kTH
1
TH 
4πk α
Another explanation

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
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

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
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.