Redshift Academy


Bosonic String Theory
---------------------

Open Strings
------------

From classical physics:

E = p²/2m + B where B is the rest energy (internal or 
binding energy).

From SR for p << c

E = √(p²c² + m²c⁴)  ... 1.

  = √(m²c⁴(1 + p²c²/m²c⁴))
 
  = mc² + p²c⁴m/2m²c⁴ from √(1 + ε) = 1 + ε/2  (binomial)
 
  = mc² + p²/2m

Light Cone Frame
----------------

To simplify matters, we want to be able to treat strings 
non-relativistically.  To accomplish this we introduce the 
LIGHT CONE FRAME.  The LCF is a 'trick' that wallows a 
relativistic system to be treated non-relativistically.

Consider a large boost along the z axis such that p_z -> c.  
Set c = 1.  



From 1.

E = √(p_z² + p_x² + p_y² + m²)

  = p_z√(1 + (p_x² + p_y² + m²)/p_z²))

  = p_z(1 + (p_x² + p_y² + m²)/2p_z²)

  = p_z + p'²/2p_z + m²/2p_z

p_z is a conserved quantity (constant).  It is the energy 
of the center of mass motion of the entire boosted string.  
It has no relevance to the internal energy of the string 
in the x, y coordinates therefore we can disregard it.  
Therefore, the energy of the string is:

E = p'²/2p_z + m²/2p_z  ... 2.

For high p_z, the energy is very small.  This is consistent 
with E = ih∂/∂t which says that the internal motions of the 
system of the system of particles are changing slowly due 
to time dilation.  The first term depends on x and y and is 
non-relativistic because the motion in the xy plane is not 
in the direction of the boost.  It is equivalent to p²/2m 
with p_z taking on the role of m.  The second term does not
depend on x and y and is equivalent to the binding or 
internal energy of the system.

Essentially, using the LCF slows the oscillations of the
string in the x, y plane.

The LCF is also called the INFINITE MOMENTUM FRAME.

Energy Stored in a String
-------------------------




FROM THIS POINT ON WE WILL ONLY CONSIDER THE X COORDS.
THERE IS A PARALLEL SET OF CALCULATIONS FOR THE Y 
COORDS.

In the x dimension:
      .
E = Σmx_i²/2 + (k/2)(Δx_i)² 

Where Δx_i = x_i - x_i+1

We make the following substitutions:

N = π/Δσ

k = N/π² = 1/πΔσ

m = μΔσ =  N/π

The energy associated with a point mass is:

E = ΣμΔσx_i²/2 + (Δσ/π)(Δx_i)² 

  = Σ(N/π)Δσx_i²/2 + (Δσ/2π)(Δx_i)² 

  = (1/2π)ΔσΣN[x_i² + (Δx_i)²]

In the continuous limit, the energy associated with 
the entire string is:
         ^π
E = (1/2π)∫(∂x/∂τ)² + (∂x/∂σ)² dσ
         ⁰
The Lagrangian is therefore:
          ^π
L = (1/2π)∫(∂x/∂τ)² - (∂x/∂σ)² dσ
          ⁰
Now, in the light cone frame (see 2.) we equated the 
internal energy with m².  Therefore,
         ^π
E = (1/2π)∫(∂x/∂τ)² + (∂x/∂σ)² dσ = m²
         ⁰
Boundary Conditions
-------------------

Now we have to consider what is happening at the string
endpoints.  There are two possible boundary conditions.  
Either the ends are fixed so they cannot move.  Or the 
ends are not fixed and are moving around in the spacetime
background.  Fixed ends are called Dirichlet boundary 
conditions.  Free ends are called Neumann boundary 
conditions.  We will see later that fixed ends are 
attached to BRANES.

Dirichlet:  x(0) = 0, x(π) = 0

Dirichlet conditions result in the reflection of waves 
at the endpoints with inversion.

Neumann: ∂x(0)/∂σ = 0, ∂x(π)/∂σ = 0

Neumann conditions result in the reflection of waves at 
the endpoints without inversion.

Now look at the boundary condition in more detail.

|
N .... N-1
|

N is the endpoint.  Force on N is F is:

F = (k/2)Δx

  = (N/2π²)Δx

  = (N/2π²)(∂x/∂σ)Δσ

  = (N/2π²)(∂x/∂σ)π/N

  = (1/2π)(∂x/∂σ)
                                    
From Newton's laws μ∂²x/∂τ² = (1/N)∂²x/∂τ².  Therefore,

(1/N)∂²x/∂τ² = (1/2π)(∂x/∂σ)

Or,

∂²x/∂τ² = (N/2π)(∂x/∂σ)

This implies that the acceleration will head towards 
infinity as N gets larger and larger.  Therefore, the 
correct boundary conditions for the open string are 
the Neumann boundary conditions.

Fourier Expansions
------------------

Strings can oscillate in any of a number of discrete 
modes.  We want to know whether the different modes 
reproduce the characteristics of the familiar particles 
of the standard model. 
                                        _∞
Fourier expansion for Neumann: x(σ,τ) = Σx_n(τ)cos(nσ)   
                                        ⁿ⁼⁰
The x_n's are the Fourier coefficients.  They depend on
the proper time, τ and, as we shall see, represent the 
harmonic oscillator modes.

Note: ∂sin(nσ)/∂σ = cos(nσ) = 0 at endpoints.
_π
∫cos(nσ)cos(mσ)dσ = 0 if m ≠ n
⁰
                  = π/2 if m = n

                  = δ_mnπ/2

                  = π if n = m = 0

Consider the classical string energy again.
          ^π
E = (1/2π)∫(∂x/∂τ)² + (∂x/∂σ)² dσ
          ⁰
              .        .
(∂x/∂τ)² = Σ_nΣ_mx_ncos(nσ)x_mcos(mσ)  [Σ is 0 to ∞]
                  ..
       = (1/2π)Σ_nΣ_mx_nx_m∫cos(nσ)cos(mσ) dσ [∫ is 0 to π]
         .              .
       = x₀²/2 + (1/2π)Σ_nx_n²(π/2)
         .             .
       = x₀²/2 + (1/4)Σ_nx_n²

(∂x/∂σ)² = Σ_nΣ_mnx_nsin(nσ)mx_msin(mσ)

        = (1/2π)Σ_nΣ_mnmx_nx_m∫sin(nσ)sin(mσ) dσ

        = (1/2π)Σ_nn²x_n²∫sin(nσ)sin(mσ) dσ

        = (1/2π)Σ_nn²x_n²(π/2)

        = (1/4)Σ_nn²x_n²

Therefore the total energy is:
    .             .
E = x₀²/2 + (1/4)Σ_nx_n² + (1/4)Σ_nn²x_n²

The first term is the KE of the centre of mass of 
the string.  The second term is the KE of the points 
oscillating in the transverse direction.  The third 
term is the PE due to tension between points on the 
string.

We can equate the string oscillation mode,n, with 
ω in the following manner:

E = hω = x₀²/2 + (1/4)Σ_nx_n² + (1/4)Σ_nn²x_n²

With h = 1 and neglecting the first term we see:

ω = (1/4)Σ_nx_n² + (1/4)Σ_nn²x_n²

String Quantization
-------------------

The second and third terms is the equation for the 
harmonic oscillator with frequency n (≡ ω_n).  These 
represent the internal energy of string identified 
with m².  We can write the Lagrangian as:
           .
L = (1/4)Σ_nx_n² - (1/4)Σ_nn²x_n²
                                               
The Hamiltonian is found from:

H = p - L 

Where,
        .   .
p = ∂L/∂x = x/2

Therefore,
 
H = p² + n²x_n²/4

  = (nx_n/2 + ip_n)(nx_n/2 - ip_n)

The commutator is:

[(nx_n/2 + ip_n),(nx_n/2 - ip_n)] = n

Therefore,

[(1/√n)(nx_n/2 + ip_n),(1/√n)(nx_n/2 - ip_n)] = 1

These correspond to the creation and annihilation 
operators that obey the commutator:

[a^-,a⁺] = 1

Now,

(nx_n/2 + ip_n) + (nx_n/2 - ip_n) = (√n)x_n

Therefore, 

x_n = (a_n^- + a_n⁺)/√n

We can write:

x(σ,τ) = Σ_nx_n(τ)cos(nσ) 

       = Σ_n((a_n^- + a_n⁺)/√n)cos(nσ)

If string theory is to be a theory of quantum gravity, 
then the average size of a string should be somewhere 
in the vicinity of the length scale of quantum gravity, 
called the Planck length, which is about 10^-33 cm.  This
is referred to as the characteristic length scale.

Particle Spin Angular Momentum
-------------------------------

# of spin states for massive particles = 2J + 1.

Spin    Massive   Massless
----    -------   --------

0          1         1
1          3         2
2          5         2

Consider a spin 1 particle.  For the massive case 
it is always possible to bring the particle to rest 
(catch up with it), rotate it and boost it in a 
direction perpendicular to the direction of motion.  
The spin can therefore be parallel, anti-parallel or 
perpendicular to the original direction of motion 
(i.e. a zero component along the direction of motion).

For the massless case it is not possible to bring the 
particle to rest and do the same thing.  Thus, the 
spin of a massless spin 1 particle is constrained to 
be parallel or anti-parallel to the direction of 
motion.  The particle either has right-handed spin 
or left-handed spin.  For the photon this manifests 
itself as the CIRCULAR POLARIZATION of light.  The 
photon can also be LINEARLY POLARIZED.  In both cases 
the polarization is always perpendicular to the direction 
of motion (transverse).  

For linear polarization, the polarization state can 
be written as:
 
|P_L> = |P_x> + |P_y>

This would be a photon linearly polarized at some 
angle to the x and y axes for a photon propagating 
in the z direction.

For circular polarization, the polarization state 
can be written as:

|P_RH> = |P_x> + i|P_y>

|P_LH> = |P_x> + i|P_y>

Note that these are similar to the ladder operators
associated with the spin angular momentum.

Discrete Energy Spectrum of an Open String
------------------------------------------

From before we have said that energy, E, corresponds 
to m².  By analogy with the ground state energy of 
the harmonic oscillator ((1/2)hω), we can assign a 
ground state mass, m₀, to the unexcited string.

a and b refer to the x and y oscillators respectively.

|0> is the ground state not the vacuum.

a^-_n|0> = 0

b^-_n|0> = 0

a⁺₁|0> = m₀² + 1 unit of energy

b⁺₁|0> = m₀² + 1 unit of energy

(a⁺₁ + b⁺₁)|0> = m₀² + 1 unit of energy

a and b are identified with coordinates x and y in the
following way:

x_n = (a_n^- + a_n⁺)/√n

y_n = (b_n^- + b_n⁺)/√n

Therefore they have vector-like properties.  In fact, they 
have similar properties to the polarization states of 
the photon.  For linear polarization we can write, for 
example.
                  
a⁺₁|0> + b⁺₁|0>

and for circular polarization.

a⁺₁|0> ± ib⁺₁|0>

This corresponds to spin angular momenta around the z 
axis of +1 and -1.

The fact that there are only spin states of +1 and -1 
and no 0 states implies that these objects could be 
PHOTONS.  If this is true, however, then they must be 
massless implying that m₀² = -1.  Unfortunately, this 
doesn't work because the ground state energy would 
then be -1.  A particle with -m₀² is called a TACHYON 
- a hypothetical particle that always moves faster 
than light.   This can be seen as follows:

E = hω = √(p²c² + m²c⁴)

Now, the group velocity, v, is defined as ∂ω/∂k (≡ ∂E/∂p)

With c = 1 we get:

∂E/∂p = p/√(p² + m²)

For positive m², v is < c.  For negative m², v is > c.  

To see how the dilemna involving Tachyons can be 
solved we need to introduce the idea of increasing 
the number of spacial dimensions in the theory.
This is discussed in a separate note.