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Bosonic String Theory

Open Strings

From classical physics:
E = p^{2}/2m + B where B is the rest energy (internal or
binding energy).
From SR for p << c
E = √(p^{2}c^{2} + m^{2}c^{4}) ... 1.
= √(m^{2}c^{4}(1 + p^{2}c^{2}/m^{2}c^{4}))
= mc^{2} + p^{2}c^{4}m/2m^{2}c^{4} from √(1 + ε) = 1 + ε/2 (binomial)
= mc^{2} + p^{2}/2m
Light Cone Frame

To simplify matters, we want to be able to treat strings
nonrelativistically. To accomplish this we introduce the
LIGHT CONE FRAME. The LCF is a 'trick' that wallows a
relativistic system to be treated nonrelativistically.
Consider a large boost along the z axis such that p_{z} > c.
Set c = 1.
From 1.
E = √(p_{z}^{2} + p_{x}^{2} + p_{y}^{2} + m^{2})
= p_{z}√(1 + (p_{x}^{2} + p_{y}^{2} + m^{2})/p_{z}^{2}))
= p_{z}(1 + (p_{x}^{2} + p_{y}^{2} + m^{2})/2p_{z}^{2})
= p_{z} + p'^{2}/2p_{z} + m^{2}/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} + m^{2}/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
nonrelativistic because the motion in the xy plane is not
in the direction of the boost. It is equivalent to p^{2}/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_{i}^{2}/2 + (k/2)(Δx_{i})^{2}
Where Δx_{i} = x_{i}  x_{i+1}
We make the following substitutions:
N = π/Δσ
k = N/π^{2} = 1/πΔσ
m = μΔσ = N/π
The energy associated with a point mass is:
E = ΣμΔσx_{i}^{2}/2 + (Δσ/π)(Δx_{i})^{2}
= Σ(N/π)Δσx_{i}^{2}/2 + (Δσ/2π)(Δx_{i})^{2}
= (1/2π)ΔσΣN[x_{i}^{2} + (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 m^{2}. Therefore,
^{π}
E = (1/2π)∫(∂x/∂τ)^{2} + (∂x/∂σ)^{2} dσ = m^{2}
^{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 .... N1

N is the endpoint. Force on N is F is:
F = (k/2)Δx
= (N/2π^{2})Δx
= (N/2π^{2})(∂x/∂σ)Δσ
= (N/2π^{2})(∂x/∂σ)π/N
= (1/2π)(∂x/∂σ)
From Newton's laws μ∂^{2}x/∂τ^{2} = (1/N)∂^{2}x/∂τ^{2}. Therefore,
(1/N)∂^{2}x/∂τ^{2} = (1/2π)(∂x/∂σ)
Or,
∂^{2}x/∂τ^{2} = (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_{mn}π/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_{n}Σ_{m}x_{n}cos(nσ)x_{m}cos(mσ) [Σ is 0 to ∞]
_{ } ._{ }.
= (1/2π)Σ_{n}Σ_{m}x_{n}x_{m}∫cos(nσ)cos(mσ) dσ [∫ is 0 to π]
. .
= x_{0}^{2}/2 + (1/2π)Σ_{n}x_{n}^{2}(π/2)
. .
= x_{0}^{2}/2 + (1/4)Σ_{n}x_{n}^{2}
(∂x/∂σ)^{2} = Σ_{n}Σ_{m}nx_{n}sin(nσ)mx_{m}sin(mσ)
_{ } = (1/2π)Σ_{n}Σ_{m}nmx_{n}x_{m}∫sin(nσ)sin(mσ) dσ
_{ } = (1/2π)Σ_{n}n^{2}x_{n}^{2}∫sin(nσ)sin(mσ) dσ
_{ } = (1/2π)Σ_{n}n^{2}x_{n}^{2}(π/2)
_{ } = (1/4)Σ_{n}n^{2}x_{n}^{2}
Therefore the total energy is:
. .
E = x_{0}^{2}/2 + (1/4)Σ_{n}x_{n}^{2} + (1/4)Σ_{n}n^{2}x_{n}^{2}
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 = x_{0}^{2}/2 + (1/4)Σ_{n}x_{n}^{2} + (1/4)Σ_{n}n^{2}x_{n}^{2}
With h = 1 and neglecting the first term we see:
ω = (1/4)Σ_{n}x_{n}^{2} + (1/4)Σ_{n}n^{2}x_{n}^{2}
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 m^{2}. We can write the Lagrangian as:
.
L = (1/4)Σ_{n}x_{n}^{2}  (1/4)Σ_{n}n^{2}x_{n}^{2}
The Hamiltonian is found from:
H = p  L
Where,
. .
p = ∂L/∂x = x/2
Therefore,
H = p^{2} + n^{2}x_{n}^{2}/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_{n}x_{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, antiparallel 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 antiparallel to the direction of
motion. The particle either has righthanded spin
or lefthanded 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}> + iP_{y}>
P_{LH}> = P_{x}> + iP_{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 m^{2}. By analogy with the ground state energy of
the harmonic oscillator ((1/2)hω), we can assign a
ground state mass, m_{0}, 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> = m_{0}^{2} + 1 unit of energy
b^{+}_{1}0> = m_{0}^{2} + 1 unit of energy
(a^{+}_{1} + b^{+}_{1})0> = m_{0}^{2} + 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 vectorlike 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 m_{0}^{2} = 1. Unfortunately, this
doesn't work because the ground state energy would
then be 1. A particle with m_{0}^{2} is called a TACHYON
 a hypothetical particle that always moves faster
than light. This can be seen as follows:
E = hω = √(p^{2}c^{2} + m^{2}c^{4})
Now, the group velocity, v, is defined as ∂ω/∂k (≡ ∂E/∂p)
With c = 1 we get:
∂E/∂p = p/√(p^{2} + m^{2})
For positive m^{2}, v is < c. For negative m^{2}, 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.