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String Theory I

GEORGE SIOPSIS AND STUDENTS

Department of Physics and Astronomy

The University of Tennessee

Knoxville, TN 37996-

U.S.A.

e-mail: [email protected]

Last update: 2006

ii

iv CONTENTS

UNIT 1

A first look at strings

Following “String Theory” by J. Polchinski, Vol.I. Notes written by students (work still in progress). For more information contact George Siopsis [email protected]

1.1 Units

First, we must explain the unit convention we are going to use. Take the fol- lowing two results from Quantum Mechanics and Special Relativity:

E = ℏω (1.1.1)

E = mc^2 (1.1.2)

These two equations link energy to frequency and mass through some con- stant of proportionality. The question is, are these constants fundamental in nature or created by man? The answer is that they are artificial creations, ex- isting purely because of the units we have chosen to work in. We could easily choose units such that ℏ = c = 1. By doing this, the number of fundamental units in the universe is reduced to 1, e.g., energy, all others being related to it:

[energy] = [1/time] = [mass] = [1/length] (1.1.3)

1.2 Why Strings?

Our motivation behind the developement of String Theory is our desire to find a unified theory of everything. One of the major obstacles that previous theories have been unable to overcome is the formation of a quantum theory of gravity, and it is in this respect that String Theory has had notable success

1.2 Why Strings? 3

This is logarithmically divergent. However, all higher-order amplitudes have the same divergence and when we sum the series in α:

Amplitude = () + α() + α^2 () + ... (1.2.7)

it yields finite expressions for physical quantities. Let us try it for the gravitational interaction between two point masses, each of mass M , separated by a distance r. The potential energy is:

V =

GM 2

r

which shows upon comparison with electromagnetism that the “charge” of gravity is eg ∼

GM. A gravitational “Hydrogen atom” will have energy levels

En ∼

E 1

n^2

, E 1 ∼

M e^4 g 2 ℏ^2

∼ G^2 M 5 (1.2.9)

Comparing with E 0 = M c^2 , we obtain the ratio

E 1 E 0

∼ G^2 M 4 ∼ e^4 g (1.2.10)

Immediately we see problems with using this charge to describe the grav- itational interaction, because eg is energy (mass) dependent. The classical scattering amplitude is

A 1 ∼ e^2 g ∼ GE^2 (1.2.11)

where E = M c^2 and the second-order contribution (exchange of two gravi- tons) is

A 2 ∼

dE′ E′^

|A 1 |^2 ∼ G^2

dE′(E′)^3 (1.2.12)

which has a quartic divergence. Worse yet, higher-order amplitudes have worse divergences, making it impossible to make any sense of the perturba- tive expansion (1.2.7) ( non-renormalizability of gravity). We may see our way to a possible solution by considering the problem of beta decay: Initially it was treated as a three body problem with the proton - neu- tron - electron interaction occuring at one vertex. When the energies of the resultant electrons did not match experiment, the theory was modified to in- clude a fourth particle, the neutrino, and the interaction was ’smeared’ out: the proton and neutron interacted at one vertex, where a W boson was cre- ated, which traveled a short distance before reaching the electron - neutrino vertex.

4 UNIT 1. A FIRST LOOK AT STRINGS

p

n

e

n

p

e

v

g g

So maybe we can solve our problems with quantum gravity by smearing out the interactions, so that the objects mediating the force are no longer point particles but extended one dimensional objects - strings.

This is the general concept from which we will proceed. It is a difficult task - the gravitational interaction must obey a much larger symmetry than Lorentz invariance, it must be invariant under completely general co-ordinate trans- formations, and we must of course still be able to describe the weak, strong and electromagnetic interactions.

In this chapter we will take a first look at strings. Initially we examine the com- pletely general equations of motion for a point particle using the method of least action, and then apply that method to the case of a general string moving in D dimensions. We will obtain the equations of motion for the string, and then attempt to quantize it and obtain its energy spectrum. This will highlight some basic results of string theory, as well as some fundamental difficulties.

1.3 Point particle

We begin by examining the case of a point particle, illustrating the method we will use for strings. The trajectory of a point particle in D-dimensional space is decribed by coordinates Xμ(τ ), where τ is a parameter of the particle’s tra- jectory. For a massive particle, τ is its proper time. X 0 will be a timelike cordi- nate, the remaining X~ spanning space. Infinitesimal distances in spacetime:

−dT 2 = ds^2 = −(dX^0 )^2 + (d X~)^2 (1.3.1)

6 UNIT 1. A FIRST LOOK AT STRINGS

We can write the action as

S = m

∫ (^) b

a

dτ

dT dτ

= m

∫ (^) b

a

dτ

dX^0 dτ

d X~ dτ

(invariant under reparametrizations τ → τ ′(τ )) from which we can define the Lagrangian for the system:

L = m

dX^0 dτ

d X~ dτ

Using the convention X˙ = dX/dτ , we write Lagrange’s equations:

d dτ

∂L

∂ X˙μ

∂L

∂Xμ^

For this Lagrangian, ∂L ∂Xμ^

and we obtain the equation of motion for the point particle:

d dτ

∂L

∂ X˙μ

d dτ

X˙μ L

To see the physical meaning of this equation, switch from τ to X 0 (or set τ = X^0 to “fix the gauge”). Then the 3-velocity is

vi^ =

dXi dX^0

and the equation of motion (1.3.8) reads

¨uμ^ = 0 , uμ^ = γ(1, ~v) , γ =

L

1 − ~v^2

i.e., that the acceleration is constant, as expected. We will now look at a better expression for the action: defining an extra field η(τ ), which at the moment is arbitrary, we write a new Lagrangian:

L =

2 η X˙μ^ X˙μ − 1 2 ηm^2 (1.3.11)

This has the nice feature that it is still valid for m = 0. Lagrange’s equation for η is d dτ

∂L

∂ η˙

∂L

∂η

1.3 Point particle 7

which implies

2 η^2

X˙μ^ X˙μ − 1 2 m

(^2) = 0 ⇒ η =

− X˙μ^ X˙μ m^2

Using this equation to eliminate η from the Lagrangian (1.3.11), we get back the original Lagrangian (1.3.5). Varying X μ^ ,we obtain from (1.3.11)

d dτ

X˙μ η

which agrees with the previous eq. (1.3.8). We will now examine the meaning of the field η(τ ). The trajectory is parame- terized by some co-ordinate of the system in terms of which an infinitesimal distance along the trajectory, ds, can be expressed. Let us suppose our trajec- tory is along the y axis. Then it would be easiest to parameterize the system with the co-ordinate y, and then ds = dy. We could, however, choose the pa- rameter to be θ, the the angle between a line drawn from a fixed point on the x axis at a distance ` to a position on the y axis. Then our new distance would be

ds = `

cos^2 θ

dθ (1.3.15)

The factor `/cos^2 θ is our η^2. It represents the geometry of the system due to our choice of co-ordinates. In Minkowski space,

ds^2 = −γτ τ dτ 2 , γτ τ = η^2 (1.3.16)

where γτ τ is the (single component) metric tensor. Under a reparametriza- tion,

τ → τ ′(τ ) , γτ τ →

dτ dτ ′

γτ τ (1.3.17)

i.e. γτ τ transforms as a tensor (this follows from the invariance of ds^2 ). We can see that the action is invariant under this transformation: we have

η′^ =

dτ dτ ′^

η (1.3.18)

and X˙μ^ transforms as

X^ ˙μ′^ = dX

μ dτ ′^

dτ dτ ′

dXμ dτ

Thus the Lagrangian transforms as

L′^ =

dτ ′ dτ

L (1.3.20)

and the action transforms as

S = m

∫ (^) b

a

dτ L = m

∫ (^) b

a

dτ ′L′^ (1.3.21)

1.3 Point particle 9

Alternatively, we may define light-cone coordinates in spacetime:

X±^ =

(X^0 ± X^1 ) X~T = (X^2 ,... , XD−^1 ) (1.3.32)

The Lagrangian reads

L =

2 η

X˙μ^ X˙μ − 1 2

ηm^2 = −

η

X˙+^ X˙−^ + 1

2 η

X~^ ˙^2

T −^

ηm^2 (1.3.33)

Let X+^ = τ play the role of time; then P+ is the Hamiltonian. X−^ and X~T are the coordinates and P− and P~T are their conjugate momenta. The Lagrangian becomes:

L = −

η

X˙−^ + 1

2 η

X˙ i^2 − 1 2

ηm^2 (1.3.34)

yielding

P− =

∂L

∂ X˙−^

η

Pi =

∂L

∂ X˙i^

η

X˙i (1.3.36)

The Hamiltonian is

P+ = X˙−P− + X˙iPi − L = P~ (^) T^2 + m^2 2 P−

Note that there is no term P+ X˙+^ because X+^ is not a dynamical variable in the gauge-fixed theory. Quantization:

[Pi , Xj^ ] = −iδji , [P− , X−] = −i (1.3.38)

Eigenstates of the Hamiltonian:

P+|k−, ~kT 〉 = ω+|k−, ~kT 〉 , ω+ =

~k T^2 + m^2 2 k−

To compare with our earlier result, define ω and k 1 by

ω+ =

(ω + k 1 ) , k− =

(ω − k 1 ) (1.3.40)

Then ω^2 − ~k^2 = ω^2 − k 12 − ~k^2 T = 2ω+k− − ~k T^2 = m^2.

10 UNIT 1. A FIRST LOOK AT STRINGS

1.4 Strings

s=0 s=l

x

x

t=

t=

Parametrize the string with σ ∈ [0, `]. In analogy to Fermat’s minimization of time, we will minimize the area of the world-sheet mapped out by the string. To find an expression for the area, pick a point on the worldsheet and draw the tangent vectors

~tτ = ∂

X~

∂τ

= X , ~~˙ tσ =

∂ X~

∂σ

= X~′^ (1.4.1)

The infinitesimal parallelepiped with sides ~tτ dτ and ~tσ dσ has area dA = |~tτ × ~tσ |dτ dσ (1.4.2) We obtain the total area by integrating over the worldsheet coordinates (τ, σ). The action to be minimized is the Nambu-Goto action

SN G = T

dA (1.4.3)

where T is a constant that makes the action dimensionless (tension of the string). Using

(~a × ~b)^2 = ~a^2 ~b^2 − (~a · ~b)^2 =

~a^2 ~a · ~b ~a · ~b ~b^2

we deduce

SN G = T

dτ dσ

− det hab , hab = ∂aXμ∂bXμ (1.4.5)

where the minus sign in the square root is because we are working in Minkowski space. hab is the two-dimensional metric induced on the worldsheet,

hab =

X~^2 X~˙ · X~′

X~^ ˙ · X~′^ X~′^2

12 UNIT 1. A FIRST LOOK AT STRINGS

The general solution to the wave equation is

Xμ^ = f (σ+) + g(σ−) (1.4.18)

where f and g are arbitrary functions. Let us write out the action explicitly with the simplifications made above. The parameter τ takes values from -∞ to +∞ and σ takes values between 0 and l, the length of the string. Anticipat- ing future results, we write the constant T

T =

2 πα′^

where α′^ is called the Regge slope. Then, from equation (81),

SN G =

4 πα′

−∞

∫ (^) l

0

dτ dσ( X˙^2 − X´^2 ) (1.4.20)

4 πα′

−∞

∫ (^) l

0

dτ dσ(∂a Xμ∂aXμ) (1.4.21)

We now examine the effects of boundary conditions which have yet to be taken into account in the equations of motion. We start off by varying the co-ordinates:

Xμ^ → Xμ^ + δXμ^ (1.4.22) Starting from equation (93), the variation in the action is

δSN G =

4 πα′

−∞

∫ (^) l

0

dτ dσ (∂aδXμ∂aXμ + ∂aXμ∂aδXμ) (1.4.23)

2 πα′

−∞

∫ (^) l

0

dτ dσ (∂aδXμ∂aXμ) (1.4.24)

And noting the total derivative

∂a(δXμ∂aXμ) = ∂aδXμ∂aXμ + δXμ∂a∂aXμ (1.4.25) The last term is just the wave equation, which equals zero, so we are left with:

δSN G =

2 πα′

−∞

∫ (^) l

0

dτ dσ∂a (δXμ∂aXμ) (1.4.26)

2 πα′

−∞

dτ

[

δXμ^ X´μ

]l 0

We will now introduce two sets of boundary conditions that will get rid of this term and leave the equations of motion unchanged at the boundary: Open string (Neumann) boundary conditions, which correspond to there be- ing no forces at the boundary:

1.4 Strings 13

X^ ´μ(σ = 0) = X´μ(σ = l) = 0 (1.4.28)

Closed string boundary conditions, which means there is no boundary and the string co-ordinates are periodic:

Xμ(σ = 0) = Xμ(σ = l) (1.4.29)

X^ ´μ(σ = 0) = X´μ(σ = l) (1.4.30)

We shall now look at deriving invariant quantities in the theory from symme- tries using Noether’s theorem. We will start with Poincare invariance, which is invariant under the transformation

Xμ^ → Λμν Xν^ + Y μ^ (1.4.31)

where Λ and Y are constant quantities. We construct the Noether current by applying this symmetry to the action. Taking the second term, we write the change in Xμ^ as δXμ^ = Y μ^ (1.4.32)

The change in the action is found by inserting this into equation (96):

δSN G =

2 πα′

−∞

∫ (^) l

0

dτ dσ∂a Y μ∂aXμ (1.4.33)

The Noether current P (^) μa is defined by

δS =

dτ dσ∂a Y μP (^) μa (1.4.34)

So in this case, P (^) μa = T ∂aXμ (1.4.35)

and ∂aP (^) μa = T ∂a∂aXμ = 0 (1.4.36)

i.e. the Poincare symmetry has led to a conserved quantity in the Noether current. Now let’s do the same for the variation

δXμ^ = Λμν Xν^ (1.4.37)

where  is a small quantity. The variation in the action is now

δSN G =

2 πα′

−∞

∫ (^) l

0

dτ dσ (Xμ∂aXν^ − Xν^ ∂aXμ) Λμν ∂a (1.4.38)

From this we can define the current as

Jaμν = T (Xμ∂aXν^ − Xν^ ∂aXμ) (1.4.39)

1.4 Strings 15

t=

y

x

t

R

We take a closed string whose initial configuration is a circle centered on the x-y origin with radius R, and whose initial velocity is ~v = 0. Then X μ^ = (t, x, y). The solution to the wave equation satisfying these boundary con- ditions is:

x = R cos

2 πτ l

cos

2 πσ l

y = R cos 2 πτ l

sin 2 πσ l

t =

2 πR l τ (1.4.47)

Let’s check the constraints X˙^2 + X´^2 = 0:

− t˙^2 + ˙x^2 + ˙y^2 − ´t^2 + ´x^2 + ´y^2 (1.4.48)

2 πR l

2 πR l

sin^2

2 πτ l

2 πR l

cos^2

2 πτ l

The total energy of the string is given by

P 0 =

∫ (^) l

0

dσP 0 τ = T

∫ (^) l

0

dσ∂τ X 0 = T

∫ (^) l

0

dσ∂τ t =

2 πRT l

∫ (^) l

0

dσ = 2πRT = E (1.4.50) The length of the string is 2 πR, so T can be identified as energy per unit length of the string, i.e. the tension.

Example 2

16 UNIT 1. A FIRST LOOK AT STRINGS

x

y

Now we consider an open string rotating in the x-y plane. The solution to the wave equation is

x = R cos πτ l

cos πσ l

y = R cos

πτ l

sin

πσ l

t =

πR l

τ (1.4.53)

The speed of each point on the string is given by

~v =

dx dt

dy dt

l πR

dx dτ

dy dτ

= cos

πσ l

− sin

πτ l

, cos

πτ l

From which we see that

~v^2 = cos^2 πσ l

Thus at the ends of the string σ = 0, l we see that ~v^2 = 1, i.e. the ends of the string travel at the speed of light. This is to be expected, since the string is massless and there are no forces on the ends of the string (Neumann boundary conditions). The intermediate points on the string don’t travel at the speed of light because they experience the tension of the string. The energy of the string is worked out exactly as before, and is found to be:

P 0 = T Rπ (1.4.56)

Let us now work out the z component of the angular momentum of the string: