# Redshift Academy

Wolfram Alpha:

Last modified: January 26, 2018
```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 pz -> c.
Set c = 1.

From 1.

E = √(pz2 + px2 + py2 + m2)

= pz√(1 + (px2 + py2 + m2)/pz2))

= pz(1 + (px2 + py2 + m2)/2pz2)

= pz + p'2/2pz + m2/2pz

pz 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'2/2pz + m2/2pz  ... 2.

For high pz, 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 pz 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 = Σmxi2/2 + (k/2)(Δxi)2

Where Δxi = xi - xi+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 = ΣμΔσxi2/2 + (Δσ/π)(Δxi)2

= Σ(N/π)Δσxi2/2 + (Δσ/2π)(Δxi)2

= (1/2π)ΔσΣN[xi2 + (Δxi)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) = Σxn(τ)cos(nσ)
n=0
The xn'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

= δmnπ/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 = ΣnΣmxncos(nσ)xmcos(mσ)  [Σ is 0 to ∞]
. .
= (1/2π)ΣnΣmxnxm∫cos(nσ)cos(mσ) dσ [∫ is 0 to π]
.              .
= x02/2 + (1/2π)Σnxn2(π/2)
.             .
= x02/2 + (1/4)Σnxn2

(∂x/∂σ)2 = ΣnΣmnxnsin(nσ)mxmsin(mσ)

= (1/2π)ΣnΣmnmxnxm∫sin(nσ)sin(mσ) dσ

= (1/2π)Σnn2xn2∫sin(nσ)sin(mσ) dσ

= (1/2π)Σnn2xn2(π/2)

= (1/4)Σnn2xn2

Therefore the total energy is:
.             .
E = x02/2 + (1/4)Σnxn2 + (1/4)Σnn2xn2

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)Σnxn2 + (1/4)Σnn2xn2

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

ω = (1/4)Σnxn2 + (1/4)Σnn2xn2

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

The second and third terms is the equation for the
harmonic oscillator with frequency n (≡ ωn).  These
represent the internal energy of string identified
with m2.  We can write the Lagrangian as:
.
L = (1/4)Σnxn2 - (1/4)Σnn2xn2

The Hamiltonian is found from:

H = p - L

Where,
.   .
p = ∂L/∂x = x/2

Therefore,

H = p2 + n2xn2/4

= (nxn/2 + ipn)(nxn/2 - ipn)

The commutator is:

[(nxn/2 + ipn),(nxn/2 - ipn)] = n

Therefore,

[(1/√n)(nxn/2 + ipn),(1/√n)(nxn/2 - ipn)] = 1

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

[a-,a+] = 1

Now,

(nxn/2 + ipn) + (nxn/2 - ipn) = (√n)xn

Therefore,

xn = (an- + an+)/√n

We can write:

x(σ,τ) = Σnxn(τ)cos(nσ)

= Σn((an- + an+)/√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:

|PL> = |Px> + |Py>

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:

|PRH> = |Px> + i|Py>

|PLH> = |Px> + i|Py>

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:

xn = (an- + an+)/√n

yn = (bn- + bn+)/√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.```