McMahon D. String Theory Demystified

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String Theory Demystiﬁ ed

The Polyakov Action

Quantization using the Nambu-Goto action is not convenient due to the presence of

the square root in the lagrangian. It is possible to write down an equivalent action,

equivalent in the sense that it leads to the same equations of motion—that does not

have the cumbersome square root. This action goes by the name of the Polyakov

action or by the more modern term the string sigma model action.

Look back to the start of the chapter when we considered the point particle. There

too, we ran into a situation where the action had a square root and we dealt with it by

introducing an auxiliary ﬁ eld

a()

. We can use the same procedure here, to rewrite the

action for the string in a more convenient form. This is done by introducing an intrinsic

metric

αβ

τσ

(, )

, which acts like the auxiliary ﬁ eld. We use the notation

αβ

because

the metric can be written as a matrix. We use the indices to denote rows and columns in

this matrix. Then, using the notation

hh= det

, the Polyakov action can be written as

dhhXX

=− − ∂ ∂

∫

ση

αβ

µν

(2.27)

A historical aside: While Polyakov did important work with this action, it was

actually proposed by Brink, Di Vecchia, and Howe and independently by Desser

and Zumino. Polyakov got his name attached to it by using it in a path integral

quantization of the string. It is also called the string sigma action.

Mathematical Aside: The Euler Characteristic

The Euler characteristic

is a number which describes the shape of a topological

space. Consider a polyhedron, and let V be the number of vertices, E be the number

of edges, and F be the number of faces. Then the Euler characteristic is

=−+VEF

(2.28)

In string theory, we often want to know whether or not two geometric shapes or

topologies are similar to one another in a speciﬁ c way. In particular, we want to

know if we can continuously deform one shape into another (imagine working with

clay and deforming one shape into another without breaking the clay apart, or

introducing or losing any holes). Formally, a homeomorphism is a deformation of a

geometric object into a new shape by stretching or compressing and being it, without

tearing or breaking it. For instance, the quintessential example is a donut and a

coffee cup (conveniently paired for police ofﬁ cers). You could use a continuous

deformation to transform one into the other or vice versa. So we say that a coffee

cup and a donut are homeomorphic. On the other hand, a sphere and a donut are not

CHAPTER 2 Equations of Motion

homomorphic—the donut has a hole but a sphere does not. The bottom line is there

is no way to transform the donut into the sphere.

If a geometric shape is homeomorphic to a sphere, then the Euler characteristic is

=−+=VEF2

(2.29)

Many shapes have an Euler characteristic which vanishes. Some examples of this

include a torus, a möbius strip, and a Klein bottle. Another example is a cylinder,

which also has

= 0

(see Fig. 2.2). Why is this interesting for us? If the worldsheet

of a string has a vanishing Euler characteristic, then it is possible to write the

auxiliary ﬁ eld

αβ

as a two-dimensional ﬂ at space metric. That is, we take [using

the choice of coordinates for the worldsheet as

(, )

τσ

]

αβ

−

⎛

⎝

⎜

⎞

⎠

⎟

(2.30)

Klein bottle

Tours

Mobius strip

linder

Figure 2.2 Some surfaces with a vanishing Euler characteristic. When the Euler

characteristic vanishes, we can deﬁ ne the auxiliary ﬁ eld such that it has

a representation of the ﬂ at space Minkowski metric.

String Theory Demystiﬁ ed

Now notice that with this choice,

hh==−det

αβ

. We also have

hXX XX XXXX

αβ

αβ ττ σσ

∂ ⋅∂ =−∂ ⋅∂ +∂ ⋅∂ =− +

′



In this case, we are able to write the Polyakov action in the remarkably simple

form

dXX

=−

′

∫

22 2

()



(2.31)

EXAMPLE 2.3

Find the equations of motion using the Polyakov action as written in Eq. (2.27)

when the auxiliary ﬁ eld takes the form of the ﬂ at space metric.

SOLUTION

In this case we have

dhhXX

dXX

=− − ∂ ∂

=− −∂ ⋅∂

∫

ση

αβ

µν

ττ

( ++∂ ⋅∂

=− − ∂ ∂ + ∂ ∂

∫

σσ

µν τ

µν σ

ση η

dXXXX

)

(

))

So, we can write the lagrangian as

LXXXX

XX X

=− ∂ ∂ + ∂ ∂

=− +

′

ηη

µν τ

µν σ

µν



µµν

′

Therefore,

∂

−+

′′

()

=− =

XX X X X



 

µµ

µν

ηη η

−−

∂

′

∂

′

−+

′′

()





XX X X

µµ

µν

ηη η

′′

′

CHAPTER 2 Equations of Motion

The Euler-Lagrange equations are

∂

⎛

⎝

⎜

⎞

⎠

⎟

+∂

∂

′

⎛

⎝

⎜

⎞

⎠

⎟



(2.32)

Hence, we ﬁ nd that the equations of motion for the relativistic string are

∂

−

∂

µµ

τσ

(2.33)

It will be convenient to call upon light-cone coordinates in string theory. First, let’s

look at how light-cone coordinates can be deﬁ ned in Minkowski space-time in

general and then consider having them in the context of the worldsheet and the

equations of motion of the string. As we will see, this will simplify the way we

write the action and the resulting equations of motion.

For simplicity, let’s take ordinary (3 + 1) dimensional space-time. The contravariant

coordinates are

xxxxx

= (,,,)

0123

where

xct

and

xxx yxz

123

===,,

say. We form light-cone coordinates by

choosing one spatial direction, which in this case we take to be x

, and forming

linear combinations of it with x

as follows:

+−

−

01 01

(2.34)

These are two null or lightlike coordinates, but you can think of x

as a timelike

coordinate and x

–

as a spacelike coordinate. Hence when we use indices and

summations, we will treat + as a “0” index and – as a “1” index. The other

coordinates x

and x

are left alone.

It is easy to derive the inverse relationship using Eq. (2.34). We have

−

+− +−

Light-Cone Coordinates

String Theory Demystiﬁ ed

Using the Minkowski metric, we have seen that inﬁ nitesimal distances in space-

time can be deﬁ ned according to

ds dx dx dx dx dx dx

2021222

=− =−−−

µν

()()()(

332

)

Since,

dx dx

−

+− +−

(2.35)

we can rewrite

in terms of light-cone coordinates as

ds dx dx dx dx

22232

2=−−

+−

()()

(2.36)

So, we can deﬁ ne distances in terms of a light-cone Minkowski metric

µν

−

⎛

⎝

⎜

⎞

⎠

⎟

0100

1000

0010

0001

(2.37)

Using Eq. (2.37), distances can be written compactly as

ds dx dx

=−

µν

(2.38)

Working with vectors is a simple extension of what we’ve written for coordinates.

That is, deﬁ ne light-cone components of a vector

+−

−

01 01

(2.39)

Using the metric from Eq. (2.37) the inner product between two vectors can be

calculated as

vw vw vw vw vw vw

⋅= = =− − +

+− −+

∑

µν

µν µ

(2.40)

CHAPTER 2 Equations of Motion

where generally,

id=−11,,…

. We can apply index raising and lowering to the

light-cone components of vectors using a sign change

vv vv

−

=− =−

The other components of the vector are left unchanged, that is,

Now that we’ve seen how to deﬁ ne light-cone coordinates for space-time, let’s

see how to deﬁ ne them for the worldsheet and hence for the string. In this case, we

deﬁ ne

στσ στσ

+−

=+ =−

(2.41)

Now since

ddd

στσ

and

ddd

στσ

−

=−

, it should be clear that

ds d d

=−

+−

σσ

(2.42)

This tells us that we can write the induced metric in Eq. (2.30) indexing a matrix as

(,,,)+++−−+−−

giving

αβ

−

⎛

⎝

⎜

⎞

⎠

⎟

012

12 0

(2.43)

You can quickly verify that the determinant is

hh==−det /

αβ

and the inverse of

Eq. (2.43) is

αβ

−

⎛

⎝

⎜

⎞

⎠

⎟

A relationship can also be written down between the derivatives with respect to the

coordinates

τσ

and the light-cone coordinates. For notational convenience, we

use the relativistic shorthand notation for derivatives

∂=

∂

and write

∂= ∂+∂ ∂= ∂−∂

+−

() ()

τσ τσ

(2.44)

String Theory Demystiﬁ ed

Let’s see how the action for the string is written using light-cone coordinates.

The Polyakov action, which we reproduce here for your convenience, is

dhhXX

=− − ∂ ∂

∫

ση

αβ

µν

Using Eq. (2.43), we ﬁ nd that

−∂∂ =− ∂∂ − ∂

+−

−+

hh X X h X X h

αβ

µν

ηη

14 14//

−−+

+−

∂

=− ∂ ∂

µν

Hence, using light-cone coordinates we ﬁ nd the Polyakov action can be written as

STd XX

=∂∂

∫

+−

ση

µν

(2.45)

We can ﬁ nd the equations of motion by varying S

. We have

δδ σ η

σδ η

µν

STd XX

Td X X

=∂∂

∫

+−

()

ddXXTdXX

σδ η σ δ η

µν

∫∫

∂∂ + ∂ ∂

+− + −

() ()

The following fact helps us proceed:

µµ

∂

±±

XX()

Therefore,

δσδησδη

µν

STd X X Td X X

=∂∂+∂∂

+− +−

∫∫

() ()

Now integrate by parts to move the derivative away from the

term. Remember

that

udv uv vdu

∫∫

=−

CHAPTER 2 Equations of Motion

In our case, we get

δσδησδη

µν

STd X X Td XX

=− ∂ ∂ − ∂ ∂

∫∫

+− −+

() ()

νν

We’ve dropped the boundary terms, which must vanish for Neumann boundary

conditions in the case of open strings or for the requirement of periodicity for closed

strings. Since

is arbitrary and

, it must be the case that

∂∂ =

+−

(2.46)

This is the wave equation for relativistic strings using light-cone coordinates.

In the next chapter, we will consider the hamiltonian and stress-energy tensor and

write down conserved charges and currents for the string. Right now, let’s focus on

ﬁ nding a solution of the wave equation given in Eq. (2.46).

From elementary mechanics, we know that the solution of a wave equation can

be written in terms of a superposition of waves moving to the left on the string and

waves moving to the right on the string. If the motion is in one dimension (call it x),

then we can write down a solution of the form

ftx fxvt fxvt

(, ) ( ) ( )=−++

We will write the equations of motion for the relativistic string in the same way.

We have a solution which is a superposition of left-moving components

τσ

()+

and right-moving components

τσ

(

)

−

XX X

µµ µ

τσ τσ τσ

(, ) ( ) ( )=++−

(2.47)

You should recall from partial differential equations that the most general solution

can be written as an expansion of Fourier modes. Here, we denote these modes as

, and write the left-moving and right-moving components as

ssk

τσ τ σ

(, ) ( )

(

=+ ++

−+

σσ

)

k≠

∑

(2.48)

ssk

τσ τ σ

(, ) ( )

(

=+ −+

−

ττσ

−

≠

∑

)

k 0

(2.49)

Solutions of the Wave Equation

String Theory Demystiﬁ ed

We have introduced some new terms here. First, we have included the

characteristic length of the string which is related to the Regge slope parameter

′

and hence to the tension in the string via

′

πα

(2.50)

Next, notice the coordinate

and momentum

. These are the center of mass

coordinate and the total momentum of the string, respectively. The “zeroth”-order

Fourier mode is deﬁ ned in terms of

αα

µµµµ

(2.51)

What does this tell us physically? The solutions imply that the string can move as a

single unit with position and momentum through space-time. In addition, it also has

vibrations, which are described by the modes

. When you see modes like this,

you should think quantization (think in terms of the harmonic oscillator or ﬁ elds in

quantum ﬁ eld theory).

Remember, we are still in the realm of classical physics, even if it’s relativistic

classical physics. So the solutions of the wave equation

XX X

µµ

,,and

must be

real functions. This implies that

and

are real (as they must be, given their

physical interpretation) and allows us to relate positive and negative modes

αα αα

µµ µµ

−−

(

)

(

)

kk kk

(2.52)

where * represents the complex conjugate. Now, let’s take a look at the solutions of

the wave equation with different boundary conditions.

Open Strings with Free Endpoints

Open strings with free endpoints satisfy the Neumann boundary condition that we

reproduce here:

∂

σπ

00when ,

CHAPTER 2 Equations of Motion

Now, looking at Eqs. (2.48) and (2.49) we see that

∂

−+

≠

∑

µµτσ

()

∂

=− −

−+

≠

∑

µµτσ

()

Summing these as in Eq. (2.47) and setting

= 0

∂

⇒= − + −

()

−

≠

∑

pp e

µµ µµ τ

αα

()

This tells us that in the case of an open string with free endpoints, it must be the

case that

µµ

()string cannot wind around itself

αα

(same modes for left- and right-moving wwaves)

Physically, this means that for an open string with free endpoints the modes

combine to form standing waves on the string. Before we move on to our next case,

let’s consider the other boundary condition, which is imposed at the other end of the

string

σπ

∂

−+

σπ

µµτπ

σσ

( ))()

≠

−+

≠

−

∑∑

−−

µµτπ

ik ik

ππ

µτ

−

≠

−

⎛

⎝

⎜

⎞

⎠

⎟

∑

()

sC iin( )k

≠

∑