**well this took me a while to write out!!**

So let me explain how this model works. First of all, it seems best to note that in most cases we are dealing with ''three neighbouring points'' on what I call a Fotini graph. Really, the graph has a different name and is usually denoted with something like [tex]E(G)[/tex] and is sometimes called the graphical tensor notation. In our phase space, we will be dealing with a finite amount of particles [tex]i[/tex] and [tex]j[/tex] but asked to keep in mind that the neighbouring particles are usually seen at a minimum three and that each particle should be seen as a configuration of spins - this configuration space is called *the spin network*. I should perhaps say, that to any point, there are two neighbours.

Of course, as I said, we have two particles in this model [tex](i,j)[/tex], probably defined by a set of interactions [tex]k \equiv (i,j)[/tex] (an approach Fotini has made in the form of on-off nodes). In my approach we simply define it with an interaction term:

[tex]V = \sum^{N-1}_{i=1} \sum^{N}_{i+1} g(r_{ij})[/tex]

I have found it customary to place a coupling constant here [tex]g[/tex] for any constant forces which may be experienced between the two distances made in a semi-metric which mathematicians often denote as [tex]r_{ij}[/tex].

If [tex]A(G)[/tex] are adjacent vertices and [tex]E(G)[/tex] is the set of edges in our phase space, (to get some idea of this space, look up casual triangulation and how particles would be laid out in such a configuration space), then

[tex](i,j) \in E(G)[/tex]

It so happens, that Fotini's approach will in fact treat [tex]E(G)[/tex] as assigning energy to a graph

[tex]E(G) = <\psi_G|H|\psi_G>[/tex]

which most will recognize as an expection value. The Fotini total state spin space is

[tex]H = \otimes \frac{N(N-1)}{2} H_{ab}[/tex]

Going back to my interaction term, the potential energy between particles [tex](i,j)[/tex] or all [tex]N[/tex]-particles due to pairwise interctions involves a minimum of [tex]\frac{N(N-1)}{2}[/tex] contributions and you will see this term in Fotini's previous yet remarkably simple equation.

[tex]K_N[/tex] is the complete graph on the [tex]N[/tex] - vertices in a Fotini Graph i.e. the graph in which there is one edge connecting every pair of vertices so there is a total of [tex]N(N-1) = 2[/tex] edges and each vertex has a degree of freedom corresponding to [tex](N-1)[/tex].

Thus we will see that to each vertex [tex]i \in A(G)[/tex] there is always an associated Hilbert space and I construct that understanding as

[tex]H_G = \otimes i \in A(G) H_i[/tex]

From here I construct a way to measure these spin states in the spin network such that we are still speaking about two particles [tex](i,j)[/tex] and by measuring the force of interaction between these two states as

[tex]F_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \hat{n}[/tex]

where the [tex]\hat{n}[/tex] is the unit length. The angle between two spins in physics can be calculated as

[tex]\mu(\hat{n} \cdot \sigma_{ij}) \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \mu(\frac{1 + cos \theta}{2})[/tex]

Thus my force equation can take into respect a single spin state, but denoted for two particles [tex](i,j)[/tex] as we have been doing, it can describe a small spin network

[tex]F_{ij} = \frac{\partial V(r_{ij})}{\partial r_{ij}} \mu(\hat{n} \cdot \sigma_{ij})^2 = \frac{\partial V(r_{ij})}{\partial r_{ij}} \mathbf{I}[/tex]

with a magnetic coefficient [tex]\mu[/tex] on the spin structure of the equation and [tex]\mathbf{I}[/tex] is the unit matrix.

I now therefore a new form of the force equation I created with an interaction term, as I came to the realization that squaring everything would yield (with our spin states)

[tex]-\frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} \mu(\hat{n} \cdot \vec{\sigma}_{ij})^2[/tex]

[tex] = -\frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} \begin{bmatrix}\ \mu(n_3) & \mu(n_{-}) \\ \mu(n_{+}) & \mu(-n_3) \end{bmatrix}^2[/tex]

Sometimes it is customary to represent the matrix in this form:

[tex]\begin{bmatrix}\ \mu(n_{3}) & \mu(n_{-}) \\ \mu(n_{+}) & \mu(-n_{3}) \end{bmatrix}[/tex]

As we have in our equation above. The entries here are just short hand notation for some mathematical tricks. Notice that there is a magnetic moment coupling on each state entry. We will soon see how you can derive the Larmor Energy from the previous equation.

Sometimes you will find spin matrices not with the magnetic moment description but with a gyromagnetic ratio, so we might have

[tex]\frac{ge}{2mc}(\hat{n} \cdot \sigma_{ij}) = \begin{bmatrix}\ g \gamma(n_3) & g \gamma(n_{-}) \\ g \gamma(n_{+}) & g \gamma(-n_3) \end{bmatrix}[/tex]

The compact form of the Larmor energy is [tex]-\mu \cdot B[/tex] and the negative term will cancel due to the negative term in my equation

[tex]-\frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} \mu(\hat{n} \cdot \vec{\sigma}_{ij})^2[/tex]

[tex]= -\frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} \begin{bmatrix}\ \mu(n_3) & \mu(n_{-}) \\ \mu(n_{+}) & \mu(-n_3) \end{bmatrix}^2[/tex]

The [tex]L \cdot S[/tex] part of the Larmor energy is in fact more or less equivalent with the spin notation expression I have been using [tex](\hat{n} \cdot \sigma_{ij})[/tex], except when we transpose this over to our own modified approach, we will be accounting for two spins.

We can swap our magnetic moment part for [tex]\frac{2\mu}{\hbar Mc^2 E}[/tex] and what we end up with is a slightly modified Larmor Energy

[tex]\Delta H_L = \frac{2\mu}{\hbar Mc^2 e} \frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} (\hat{n}\cdot \sigma_{ij}) \begin{pmatrix} \alpha \\ \beta \end{pmatrix}[/tex]

This is madness I can hear people shout? In the Larmor energy equation, we don't have [tex](\hat{n}\cdot \sigma) \begin{pmatrix} \alpha \\ \beta \end{pmatrix}[/tex] we usually have [tex](L\cdot S)[/tex]?

Well yes, this is true, but we are noticing something special. You see, [tex](L\cdot S)[/tex] is really

[tex]|L| |S|cos \theta[/tex]

This is the angle between two vectors. What is [tex](\hat{n}\cdot \sigma) \begin{pmatrix} \alpha \\ \beta \end{pmatrix}[/tex] again? We know this, it calculates the angle between two spin vectors again as

[tex]\frac{1 + cos \theta}{2}[/tex]

So by my reckoning, this seems perfectly a consistent approach.

Now that we have derived this relationship, it adds some texture to the original equations. If we return to the force equation, one might want to plug in some position operators in there - so we may describe how far particles are from each other by calculating the force of interaction - but as we shall see soon, if the lengths of the triangulation between particles are all zero, then this must imply the same space state, or position state for all your [tex]N[/tex]-particle system. We will use a special type of uncertainty principle to denote this, called the triangle inequality which speaks about the space between particles.

As distances reduce between particles, our interaction term becomes stronger as well, the force between particles is at cost of extra energy being required. Indeed, for two particles [tex](i,j)[/tex] to experience the same position [tex]x[/tex] requires a massive amount of energy, perhaps something on the scale of the Planck Energy, but I have not calculated this.

In general, most fundamental interactions do not come from great distance and focus to the same point, or along the same trajectories. This actually has a special name, called Liouville's Theorem. Of course, particles can be created from a point, this is a different scenario. Indeed, in this work I am attempting to built a picture which requires just that, the gradual seperation of particles from a single point by a vacua appearing between them, forced by a general instability caused by the uncertainty principle in our phase space.

As I have mentioned before, we may measure the gradual seperation of particles using the Lyapunov Exponential which is given as

[tex]\lambda = \epsilon e^{\Delta t}[/tex]

and for previously attached systems eminating from the same system, we may even speculate importance for the correlation function

[tex]<\phi_i, \phi_j> = e^{-mD}[/tex]

where [tex]D[/tex] calculates the distance. Indeed, you may even see the graphical energy in terms maybe of the Ising model which measures the background energy to the spin state [tex]\sigma_0[/tex] - actually said more correctly, the background energy

[tex]\sum_N \sigma_{(1,2,3...)}[/tex]

acts as coefficient of sigma zero. Thus the energy is represented by a Hamiltonian of spin states

[tex]H = \sigma(i)\sigma(j)[/tex]

Now, moving onto the implications of the uncertainty principle in our triple intersected phase space (with adjacent edges sometimes given as [tex](p,q,r)[/tex], there is a restriction that [tex](p+q+r)[/tex] is even and none is larger than the sum of the other two. A simpler way of trying to explain this inequality is by stating: [tex]a[/tex] must be less than or equal to [tex]b+c[/tex], [tex]b[/tex] less than or equal to [tex]a+c[/tex], and [tex]c[/tex] less than or equal to [tex]a+b[/tex].

It actually turns out that this is really a basic tensor algebra relationship of the irreducible representions of [tex]SL(2,C)[/tex] according to Smolin. If each length of each point is necesserily zero, then we must admit some uncertainty (an infinite degree of uncertainty) unless some spacetime appeared appeared between each point. Indeed, because each particle at the very first instant of creation was occupied in the same space, we may presume the initial conditions of BB were highly unstable. This is true within the high temperature range and can be justified by applying a strong force of interactions in my force equation. The triangle inequality is at the heart of spin networks and current quantum gravity theory.

For spins that do not commute ie, they display antisymmetric properties, there could be a number of ways of describing this with some traditional mathematics. One way will be shown soon.

Spin has close relationships with antisymmetric mathematical properties. An interesting way to describe the antisymmetric properties between two spins in the form of pauli matrices attached to particles [tex]i[/tex] and [tex]j[/tex] we can describe it as an action on a pair of vectors, taking into assumption the vectors in question are spin vectors.

This is actually a map, taking the form of

[tex]T_x M \times T_x M \rightarrow R[/tex]

This is amap of an action on a pair of vectors. In our case, we will arbitrarily chose these two to be Eigenvectors, derived from studying spin along a certain axis. In this case, our eigenvectors will be along the [tex]x[/tex] and [tex]z[/tex] axes which will always yield the corresponding spin operator.

[tex](d \theta \wedge d\phi)(\psi^{+x}_{i}, \psi^{+z}_{j})[/tex]

with an abuse of notation in my eigenvectors.

It is a 2-form (or bivector) which results in

[tex]=d\theta(\sigma_i)d\phi(\sigma_j) - d\phi(\sigma_j)d\phi(\sigma_i)[/tex]

This is a result where [tex]\sigma_i[/tex] and [tex]\sigma_j[/tex] do not commute.