# A New Realization Into The Second Order Dirac Operator And Planck Particles

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### #1 Aethelwulf

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Posted 11 October 2012 - 06:59 PM

Planck Particles

The Concept of Planck Particles

Planck particles are very interesting subatomic systems which is generally regarded as a type of miniature black hole. This kind of black hole only exist for very small amounts of time, but the physics behind such an exotic object are interesting to say the least. A Planck particles wavelength is usually set equal or approximated to its Schwarzschild radius. It would be a very small particle indeed, with a very large mass on the scale of the Planck Mass.

We can obtain the relationship directly between the wavelength and the Scwarzschild radius directly by inferring first on a very special quantization condition which is given as

$\hbar c = GM^2$

This is a quantization of charge and one may see this because of the Heaviside relationship

$e = \sqrt{4 \pi \epsilon \hbar c}$

Because $\hbar c$ is set equal to $GM^2$ I determined that this relationship is also true

$e = \sqrt{4 \pi \epsilon GM^2}$

where $\sqrt{G}M$ is the gravitational charge. We often see the mass as the Planck mass so the reader must keep this in mind. Now, one can derive very easily the relationship for the Planck particles wavelength and it's Schwarzschild radius by dividing the inertial energy by the quantization condition as

$\frac{\hbar c}{Mc^2} = \frac{GM^2}{Mc^2} = (\frac{\hbar}{Mc} = \frac{2GM}{c^2} = r_s)$

where $r_s$ is the Scwarzschild radius and the factor of 2 comes from the usual convention for it.

The Work of Lloyd Motz

In Motz' work (1), he set the Guassian curvature equal to the radius by equation

$8 \pi \rho (\frac{G}{c^2}) = 6(\frac{\hbar}{Mc})^{-2}$

The Gaussian curvature of a 2D surface is the product of two principle curvatures. Each principle curvature is equal to $1/R$, the Gaussian curvature has dimensions $1/ \ell^2$ and the three dimensional case of a hypersphere is $K/6 = R^{-2}$ which is how these extra factors come up where $R$ is the radius of curvature.

A Gravitational Energy

When considering a gravitational charge $\sqrt{G}M$ one may consider also a gravitional energy which can be set equal to this with an extra term of the Schwarzschild radius

$\sqrt{E_gr_s} = \sqrt{G}M$

I call it a gravitational energy (over an inertial energy) because we are talking about a gravitational charge on our system. However, in all simplistic sense of the words, inertial quantities and gravitational quantities are the same thing. These are ''intrinsic properties'' in this work. A photon may exert gravitational influences by curving spacetime around it, but it contains no instrinsic gravitational charges, nor does it intrinsically relate to any gravitational energies. This new distinction helps solve the problem highlighted in this paper: http://www.tardyon.d...hooft/hooft.htm

In this paper, the authors argue there is a problem concerning what we consider as mass and a system which exerts a gravitational influence on the surrounding vicinity of the spacetime vacuum. I show, that by discerning a difference when talking about intrinsic properties, this problem can be easily avoided.

Interestingly, if you make the radius go to zero in this equation

$\lim_{r_s \rightarrow 0} \sqrt{E_gr_s} = \sqrt{G}M = 0$

then the gravitational charge (and thus), the gravitational energy (in the intrinsic sense is zero). Obviously then, we don't mean that an object without a gravitational energy cannot exert gravitational forces which stays strictly within the predictions of relativity.

Because of the relationship then:

$e^2 = 4\pi \epsilon_0 GM^2 = 4 \pi \epsilon_0 \hbar c$

we can also see that the elementary charge itself is equal to

$e = \sqrt{4 \pi \epsilon E_g r_s}$ *

which is an entirely new relationship derived for this paper. It is interpretated as saying, that the gravitational energy covers the length of the radius, which defines the charge of your system.

Why this might be interesting is because there is no such thing as a charged massless system. The Yang Mills Equations once predicted massless charge bosons but the approach has been generally considered today as a false one. Is this failure important? Is there an intrinsic relationship between the elementary charge and what we might call the inertial mass ?

The above equation, given by the asterisk *, is to be taken to mean that the elementary charge $e$ is something associated to the gravitational energy $E_g$ in terms of the gravitational charge $\sqrt{G}M$ - if we are dealing with classical sphere's, the idea is that the gravitational energy would take on this value inside the sphere of radius . Of course, the charge itself is also distributed over such a radius.

Indeed, if the Scwartzschild radius goes to zero again, then so does the gravitational energy and thus the elementary charge must also vanish. There are such cases in nature where this might be an important phase transition. Gamma-gamma interactions can either give up mass in the form of special types of decay processes or, vice versa, the energy can come from antiparticle interactions. This is called the parapositronium decay. Such a phase transition is given as

$\gamma \gamma \rightarrow e^{-}e^{+}$

and

$e^{-}e^{+} \rightarrow \gamma \gamma$

This relationship is best seen in light of my equation

$\lim_{r_s \rightarrow 0} \sqrt{E_gr_s} = \sqrt{G}M = 0$

The radius is concerned with bodies of mass. If the radius shrinks to zero, then what we have is a similar transition phase between systems with mass and massless systems. Therefore, gravitational energy is obtained through the presence of the Scwarzschild radius. The fundamental reasons behind mass, of course, will be much more complicated. This is only an insight into mass in terms of the radius itself and gravitational energy when in relation to the gravitational charge (or inertial mass) of the system.

The Origin of the Quantization of Charges

Here is a nice excerpt about the charge quantization method adopted by Motz

http://encyclopedia2....tion condition

What we have is

$e \mu = \frac{1}{2} n \hbar c$

where your constants in the paper have been set to natural units and the angular momentum component comes in multiples of $n$.

It seems to say that $\mu$ plays the role of a magnetic charge - this basically squares the charge on the left handside, by a quick analysis of the dimensions previously analysed.

Here, referenced by Motz, you can see the magnetic charge is given as $g$, the electric charge of course, still given by $e$.

http://wdxy.hubu.edu....0535154942.pdf

what we have essentially is

$\frac{e\mu - e \mu}{4\pi} = n \hbar c$

In light of this, one may also see this can be derived from the Heaviside relationship since it has the familiar $4 \pi$ in it.

$\hbar = RMc$

Knowing the quantized condition $\hbar c = GM^2$ we may replace $\hbar$ for

$\frac{GM^2}{c} = RMc$

Rearranging and the solving for the mass gives

$\frac{Rc^2}{G} = M$

we may replace the mass with the definition of the Planck mass in this equation, this gives

$\frac{Rc}{G} = \sqrt{\frac{\pi \hbar c}{G}}$

Actually, this is not the Planck mass exactly, it is about a factor of $\pi$ greater, however, the Planck mass is usually an approximation so what we have here is possibly the correct value.

After rearraging and a little further solving for the radius we end up with

$R = \sqrt{\frac{\pi G \hbar}{c^2}}$

which makes the radius the Compton wavelength. In a sense, one can think of the Compton wavelength then as the ''size of a particle,'' but only loosely speaking.

Motz' Uncertainty

And so, I feel the need to explain an uncertainty relationship Motz has detailed in paper. But whilst doing so, I also feel the need to explain that such a black hole particle is expected to give up its mass in a form of Unruh-Hawking radiation. The hotter a black hole is, the faster it gives up it's radiation. This is described by the temperature equation for a black hole

$T \propto \frac{hc^3}{4\pi kGM}$

The time in which such a particle would give up the energy proportional to the temperature is given as

$t \propto \frac{\hbar G}{c^5}$

That is very quick indeed, no experimental possibilities today could measure such an action. Uncertainty in both energy and time is given as

$\Delta E \Delta t \propto \hbar$

Motz explains that the smallest uncertainty in

$RMc \leq \hbar$

is

$\frac{\hbar G}{c^3}M^2c^2 \leq \hbar^2$

This leads back to the quantization condition, Motz explains

$\hbar c = GM^2$

Now, I came to realize the uncertainty principle is actually related to the gravitational charge.

If you consider the shortest observation of any possible system in physics exists within the so-called Planck Time. The Planck length has units of

$\ell = ct$

If $t$ is measures in seconds, then $\ell$ is measures in meters, then $c = 3 \times 10^{8}$. To measure it we need a clock with an uncertainty of $\Delta t$ can be no larger than $t$. Time and energy uncertainty says the product of $\Delta t$ and $\Delta E$ can be no less than $\frac{\hbar}{2}$

$\Delta E t \leq \Delta E \Delta t \leq \frac{\hbar}{2}$

so that

$\Delta E \leq \frac{\hbar}{2t} \leq \frac{\hbar c}{2 \ell}$

which implies a relationship

$\Delta E \leq \frac{GM^2}{2\ell}$

The temperature of a system is related also to the gravitational charge as

$\frac{GM^2}{4\pi r_s} = kT$

Solving for the gravitational charge we end up with

$\frac{4\pi \hbar}{Mc} kT = GM^2$

Where $k$ is the Boltzmann constant. Plugging in the definition of the Planck Temperature, we find that

$\frac{4 \pi \hbar}{Mc} k\sqrt{\frac{\hbar c^5}{Gk^2}} = \frac{4 \pi \hbar}{Mc}k \frac{Mc^2}{k}$

Brings us also back to the gravitational charge definition

$\hbar c = GM^2$

Which is more important than first meets the eye.

An Ambiguity of the Black Hole Charge

There does exist, a certain ambiguity for those who are familiar with the Black Hole charge I wish to discuss - that being, why is the gravitational charge this, and not the integral of taken over a two dimensional surface defining the gravitational charge as

$M_g = \frac{1}{4\pi} \int g ds$

Well, the reason has to do with the quantization condition. The equation above is true for your usual standard black hole (your very large objects), but to describe Planck Particles, you would need work with the quantum relationships. This can be understood if we (just for now) defined the gravitational charge as $\mu$ and saw that

$4\pi \frac{\epsilon \mu}{Mc^2} = \ell_P$

is in fact the gravitational analogue of

$\frac{e^2}{Mc^2} = R_{classical}$

which defines the classical electron radius, so the squared gravitational charge certainly has it's right place.

It's not taken in this paper that the classical electron radius is the exact size of the particles we are dealing with, only that we wish to keep an open mind that particles are not truly pointlike. It is determined that they are pointlike down to scales of $10^{-18}$m, which does not mean that particles are actually pointlike, only that that electrons for example, behave as though they are a single object with a $1/r$ potential without any extra degrees of freedom. (2)

The CODATA charge

In the CODATA method of understanding charge, the idea is simply this: treating charge not as a independent quantity, but rather but a relationship of fundamental constants.

This is a wise move, since we are often taught that the relationships in nature are not by accident, that there may be some fundamental written set of rules which determine charges for systems. Because of this, we may understand that perhaps the gravitational charge is also strictly governed by similar principles.

The charge due to fundamental relationships is given as

$e = \sqrt{\frac{2\alpha \pi \hbar}{\mu_0 c}}$

Therefore, because of the relationship we have already covered:

$e^2 = 4\pi \epsilon_0 GM^2 = 4 \pi \epsilon_0 \hbar c$

by plugging in this definition of the elementary charge for the gravitational charge expression, and solving for the gravitational we end up at a relationship

$\sqrt{G}M = \sqrt{\frac{\alpha \hbar}{2 \epsilon_0 \mu_0 c}}$

But this is not where this speculation ends. The quantity $\sqrt{\alpha \hbar}$ is actually quite important. A curious interpretation of $e^2 = \alpha \hbar c$ is that the angular momentum component is in fact conserved by the fine structure constant by stating

$\frac{e^2}{c} = \pm \alpha \hbar n$

this means in my equation we actually have

$\sqrt{G}M = \sqrt{\frac{\pm \alpha \hbar n}{2 \epsilon_0 \mu_0 c}}$

this means the gravitational charge itself can be understood as depending on the conservation of angular momentum as well. This was quite an important discovery, I felt.

Two Famous Relationships

Barrow and Tipler calculated a type of gravitational charge in their equations, when considering the ratio of an electric charge with the gravitational charge:

$\frac{\alpha}{\alpha_G} = \frac{e^2}{GM_pM_e} = \frac{e^2}{GM^2}$

The fine structure constant can also be given as

$\alpha_G = \frac{\hbar c}{GM^2} = (t_p \omega)^2$

Coupling of the Gravielectric and Gravimagnetic Fields with the Gravitational Charge

The Coriolis Field

The Coriolis field is a gravimagnetic field. Rotating objects (or even those with intrinsic angular momentum) couple to this field.

Now going back to Motz paper, which is referenced on page 1, post 1, but here it is again: http://www.gravityre...d/1971/motz.pdf Motz explains how there is a coupling of matter to the gravimagnetic field through the effort of a cross product, which he explains is fully discussed by Sciarma. Now I went to some effort to find Sciarma's paper as well:

It's free which is good. Motz explains how a moving charge $\sqrt{G}M$ with velocity $v$ is not only coupled to Newtonian gravitational fields and gravielectric fields, but also to the Coriolis Field, which is the gravimagnetic field which he has defined as $\frac{2\omega c}{\sqrt{G}}$. This seems to have come from (my guess) that the Coriolis acceleration is

$a = 2(\omega \times v)$

and the force is

$F = -2M\omega \times v$

so you must get his quantity by dividing the force by the gravitational charge

$\frac{F}{\sqrt{G}M} = \frac{2\omega v}{\sqrt{G}}$

Such a coupling of the charge to gravimagnetic field is achieved, as explained by Motz

$\sqrt{G}M \frac{v}{c} \times \frac{2\omega c}{\sqrt{G}}$

according to Motz. Now, I will explain something else. The cross product of the terms $v \times \omega$ actually give rise to a matrix determinent, which I am not going to write out but I hope you can take my word for it. The Lorentz force is of course $evB$ and to refresh our minds, the Coriolis force is $-2M\omega v$ (where we are omitting the cross products). What is interesting is if you set them equal,

$evB = -2M\omega v$

(setting these two quantities equal with each other should not be a surprise, since the Coriolis force is a type of gravimagnetic field)

cancel the linear velocities and divide the gravitational charge on both sides you get

$B = -\frac{2 \omega}{\sqrt{G}}$

Now, according to Sciarma, the gravimagnetic field is in general not zero and I also extend this idea, that there has been no fundamenta particles in nature which have been found with zero spin, (maybe with exception one). As of recently, there has been speculations that we might have found a Higgs Boson due to a certain energy signature. Nothing is yet certain however and it is also sepculated that it possibly isn't a Higgs Boson because it does not have the same energy requirements.

Rotating Sphere's

As I have explained, I don't believe particles are truly pointlike. One reason lyes in the same objection raised by the paper in reference 2, that being that particles with zero radii like an electron for instance, result in infinities energies due to their bare mass. Renormalization proceedures are often adopted to try and solve the problem, but there has been some speculation over the years whether renormalization is even the correct approach. One example is Paul Dirac who was often voiced on the subject.

I begin by referencing a well-known equation

$G = \frac{rc^2}{2Gt^2}$

From this equation, one can derive this relationship:

$\frac{GM^2}{\hbar} = (\omega \times r_s)$

The derivation is quite long, so I won't be writing it out for the purpose of getting through this as quick as possible. One can take my equation

$B = -\frac{2\omega}{\sqrt{G}}$

and do the cross product with the term on the right of the equation just submitted. We have

$-\frac{2\omega}{\sqrt{G}} \times (\omega \times r_s) = \frac{\omega \times v}{\sqrt{G}}$

Where on the left, we can see Motz' term he defines as the gravimagnetic field. One can obtain this by saying

$ev \times B = M\omega \times v$

and divide the gravitational charge on both sides we end up

$v \times B = \frac{\omega \times v}{\sqrt{G}}$

and thus by substituting all the respective terms together, one can end up with this relationship:

$-\frac{2\omega}{\sqrt{G}} \times (\omega \times r_s) = B \times v$

Interestingly, $(\omega \times r_s)$ is just a rotational velocity. It certainly seems appropriate to consider rotating bodies coupling to such gravimagnetic fields. $$\omega$$ is actually perpendicular to the radius component $\omega \perp r_s$. However, where we have speculated a rotational velocity, this only applies to systems which are not pointlike but rather sphere's. Sphere's of course will have rotational velocities. In this paper, I don't speculate the size of the sphere's we may be dealing with, only that we are dealing with them in some of these equations.

Similarities to Gravitational and Electromagnetic Forces

It perhaps would have been a good idea to cover a little more than I had done, concerning the analogies of electric field equations and those concerned with the gravitational.

Let's choose a few. Motz defines a few of his own:

$F = \frac{\sqrt{G}M_1 \sqrt{G}M_2}{r^2}$

This is of course analogous to Coulomb's Force $\frac{q_1q_2}{r^2}$.

He notes, that the gravitational charge will be the source of the gravitational field

$\frac{\sqrt{G}M}{r^2}$

If anyone has noticed, this is actually analogy of the magnitude of the electric field $\mathbb{E}$ which is created by a single charge

$\mathbb{E} = \frac{1}{4\pi \epsilon} \frac{q}{r^2}$

Another one to note perhaps, would be the appearance of the gravimagnetic field, given by Motz as $\frac{-2\omega v}{\sqrt{G}}$. We derived what this was equated to in the second post, it is actually equated to

$\frac{F}{\sqrt{G}M} = -\frac{2\omega v}{\sqrt{G}}$

This is actually the gravitational analogue of the equation

$\mathbb{E} = \frac{F}{q}$

which describes the definition of the electric field. So the gravitational case, must be the definition of the gravimagnetic field or something akin to it since

$-\frac{2\omega v}{\sqrt{G}}$

is also

$v \times B$

we determined not so long ago.

The Final Proposed Equation to Measure Spinning Black Hole Particles

And so this brings me to the end and an equation which will help bridge the gap of Motz' work extended to Planck Particles. The equation proposed is

$-\frac{2\omega}{\sqrt{G}} \times (\omega \times 8 \pi \rho_0 (\frac{G}{c^2})) = B \times v$

Where $8 \pi \rho_0 (\frac{G}{c^2})$ is the Gaussian curvature, $\rho_0$ is the proper density and $\frac{G}{c^2}$ is the Schwarzschild constant which from now on, we will define it by a new constant, given as $S_c$. This new equation helps describe the relationship with the magnetic field cross product with velocity as being related to magnetic field obtained (which was obtained by dividing by the gravitational charge on the both sides of a previous equation) as a cross product with the angular velocity itself further cross product with the curvature of the particle.

The justification for this, is that we have already obtained that in Motz work, he has set the Compton wavelength equal to the Gaussian curvature. But since we are talking about Planck particles, this means that the radius of curvature must itself be the same thing as the Schwarzschild radius $r_s = R$ where $R$ is the radius of curvature. This means that $r_s$ can be easily seen to be interchangeable by definition in my equation.

Invariant Time Operator and Temperature as a Function of the Planck Time

We present this form of the Uncertainty Principle

$\Delta \frac{GM^{2}_{p}}{\ell_p} \Delta t_p \propto k\Delta T(t_p) \frac{R}{c}$

The $R$ in the operator can be taken as either the Schwarzschild radius, or the Compton Wavelength.

The temperature depends on the time and this can be understood because large black holes are cold, whilst the smaller a black hole is much hotter, it radiates faster and so depends on the time which passes by.

Spinors and the Dirac Operator

This continued work will concern the more complicated work of Spinors, with two spin matrices and the Dirac operator. Motz was influential massively with the work I calculated. First of all, the second order Dirac Operator for spinors is

$D \psi(x) = \hbar^2 R^{-2} \psi(x)$

This was the way Motz presented it, but I am going modify the Planckian constant and re-write this for a two dimensional case and also rewrite the equation in terms of it's matrix components and also treating the spin with a special semi metric notation. Basically do to so, we much recognize that there are two particles being particle $\pm \sigma_{(i,j)}$.

Now as I explained, Planck Particles cannot be massless and there is some evidence that are actually fermionic by nature meaning they will obey the Dirac Equation, which means that will also be subjected to spin-up, spin-down Eigenvalues.

Writing the two dimensional wave function for the Dirac Operator in terms of Spinors, we obtain

$\psi(x,y) = \begin{pmatrix} \chi & (x,y) \\ \eta & (x,y) \end{pmatrix} |i^{\uparrow}, j_{\downarrow}) \in \mathcal{R}^{2}$

For a three dimensional case, involves

$D = \gamma^{\mu} \partial_{\mu}$ which can also be written in a Feymann slash notation.

The Spin of a Dirac Operator is therefore provided as

$D = -i \sigma_x \partial_x - i \sigma_y \partial_y$

This is quite important because it implies anticommutation between the spin states for our particles. I can quickly explain how...

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 $i$ and $j$ 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

$T_x M \times T_x M \rightarrow R$

This is a map 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 $x$ and $y$ axes which will always yield the corresponding spin operator.

$(d\theta \wedge d\phi)(\psi^{+x}_{i}, \psi^{+z}_{j})$

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

$=d\theta(\sigma_i)d\phi(\sigma_j) - d\theta(\sigma_j)d\phi(\sigma_i)$

Motz and the Invariant Time Operator

Taking the general gist of his argument, $R$ the curvature can be more understood in terms of the second order Dirac operator, namely

$D \psi(x)= \hbar^2 R^{-2}\psi(x)$

If one factors this equation in terms of the gamma matrices four tuple $\gamma^{\mu}$ we obtain $i\gamma^{\mu}\nabla_{\mu}R = 1$ from which Motz deduces that $R$ must be the squared spacetime interval.

Now as I explained, for a Planck Particle, one must set the Compton wavelength equal to the Schwarzschild radius. This means that the Gaussian curvature equation can be rewritten in this form

$8 \pi \rho (\frac{G}{c^2}) = (\frac{12GM}{c^2})^{-2}$

(note this condition does not hold for Motz' outdated Uniton particle)

The factor of $12$ arises because there is a coefficient of $2$ on the gravitational parameter which makes the Schwarzschild radius. Since this simply has dimensions of $R^{-2}$ then this means we can rewrite the Dirac Operator as

$D \psi(x)= \hbar^2 (\frac{12GM}{c^2})^{-2}\psi(x)$

The gravitational charge must be understood then as the coefficient of the radius of the curvature of our system, since the quantization condition is

$\hbar = \frac{GM^2}{c}$

If we use my energy-radius relationship which describes the gravitational charge and replace it in the numerator of the quantization condition we have

$\hbar = \frac{Er_s}{c}$

Let us set $r_s = R$ then place it into the new Dirac Operator:

$D \psi(x)= E^2\frac{R^2}{c^2} (\frac{12GM}{c^2})^{-2}\psi(x)$

Where $(\frac{R}{c})^2$ plays the invariant time operator role squared.

Finishing up and the Final Conclusions

Let us go back to the modified second order Dirac Operator:

$D \psi(x)= E^2\frac{R^2}{c^2} (\frac{12GM}{c^2})^{-2}\psi(x)$

So... why is it important that $(\frac{R}{c})^2$ plays the invariant time operator role? Keep in mind first of all, I define the gravitational charge as

$E_gr_s = GM^2$

as the energy itself multiplied by the systems Schwarzschild radius. Whilst many may not think this is an important relationship, it does explain itself how the energy could contribute to the inertia of system, just as Einstein himself speculated.

In Motz' work, he defines the well-known quantum relationship

$\hbar = RMc$

The Gaussian curvature $K$ has the form $\frac{K}{6} = (\frac{\hbar}{mc})^{-2}$. Then Motz introduces the curvature of radius for a particle $R$ for a three dimensional hypersphere as

$K/6 = R^{-2}$

from which one can obtain

$R^2 = (\frac{\hbar}{Mc})^2$

or simply from the original relationship

$\hbar = RMc$

as he mentions. He notes that one may look at this as an uncertainty relationship, between the rest energy $Mc^2$ and the radius of the particle... Or put better, the space-interval it occupies. He determines that this must mean that the radius can be represented by a dynamical operator, that does not commute with the mass term $M$. Therefore the equation he proposed was

$\frac{R}{c}Mc^2 = \hbar$

...in which $\frac{R}{c}$ he determines is the time operator.

Now his speculations are beginning to take shape, except understood perhaps in better ways when one realizes the consequences of Planck Particles, and how the invariant time operator plays it's part in the modified form of the second order Dirac Operator

$D \psi(x)= E^2\frac{R^2}{c^2} (\frac{12GM}{c^2})^{-2}\psi(x)$

We can easily interpret that the coefficient on the operator (the energy squared) as being an inertial energy, since this is what the work I have been speculating on is all about... that being that the gravitational charge is all about it's inertial content, with the product of it's radius. Of course, Motz was careful when he mentioned the inertial momentum [*] as being the mass which is what is concerned with here, as much as the mass term in the modified Dirac equation has a mass term equally.

[*] However, I noticed a mistake in his paper and his inertial momentum condition wouldn't hold because it had the wrong dimensions, but could be fixed as an inertial energy. It is possible he simply got confused when writing it.

Conclusions

Now... would would that mean?

Well it would mean there is equally an uncertainty relationship in the modified second order Dirac Operator, when you consider Planck Particles, rather than the path Motz took, using the idea of Unitons to explain the fundamental nature of particles. This also means the more you tried to locate the radius of a Planck Particles mass, the more uncertain it's mass will become and vice versa.

So the picture I have presented, begins to get easier when we realize, we are dealing within Planck Parameters, or Scales as they are usually called. The smallest uncertainty in the radius (which would be the Planck Length) must be approximated to the uncertainty in the Planck Mass, which is actually very large. The thing about this though, is that this could be the reason why such particles are so fundamentally unstable - it is true we understand it in terms of the temperature of the particles, and I even gave many demonstrations of this... in my previous work. But what is the real fundamental reason for it's unstable nature? Is it because of this apparent inherent uncertainty principle between the radius and the rest mass?

I think it's just as important as the realization that the uncertainty in the temperature leads back the quantization condition $\hbar c = GM^2$, as shown previously. It seems ... too much of a coincidence to just pass to the side so easily.

As I said before...

Invariant Time Operator and Temperature as a Function of the Planck Time

We present this form of the Uncertainty Principle

$\Delta \frac{GM^{2}_{p}}{\ell_p} \Delta t_p \propto k\Delta T(t_p) \frac{R}{c}$

The $R$ in the operator can be taken as either the Schwarzschild radius, or Compton wavelength.

The temperature depends on the time and this can be understood because large black holes are cold, whilst the smaller a black hole is much hotter, it radiates faster and so depends on the time which passes by.

and in respect of what I said about the temperature leading back to the gravitational, just to refresh the mind

Now, I came to realize the uncertainty principle is actually related to the gravitational charge.

If you consider the shortest observation of any possible system in physics exists within the so-called Planck Time. The Planck length has units of

$\ell = ct$

If $t$ is measures in seconds, then $\ell$ is measures in meters, then $c = 3 \times 10^{8}$. To measure it we need a clock with an uncertainty of $\Delta t$ can be no larger than $t$. Time and energy uncertainty says the product of $\Delta t$ and $\Delta E$ can be no less than $\frac{\hbar}{2}$

$\Delta E t \leq \Delta E \Delta t \leq \frac{\hbar}{2}$

so that

$\Delta E \leq \frac{\hbar}{2t} \leq \frac{\hbar c}{2 \ell}$

which implies a relationship

$\Delta E \leq \frac{GM^2}{2\ell}$

The temperature of a system is related also to the gravitational charge as

$\frac{GM^2}{4\pi r_s} = kT$

Solving for the gravitational charge we end up with

$\frac{4\pi \hbar}{Mc} kT = GM^2$

Where $k$ is the Boltzmann constant. Plugging in the definition of the Planck Temperature, we find that

$\frac{4 \pi \hbar}{Mc} k\sqrt{\frac{\hbar c^5}{Gk^2}} = \frac{4 \pi \hbar}{Mc}k \frac{Mc^2}{k}$

Brings us also back to the gravitational charge definition

$\hbar c = GM^2$

(1) http://www.gravityre...d/1971/motz.pdf

(2) http://www.cybsoc.org/electron.pdf

(3) http://arxiv.org/ftp...9/1109.3624.pdf (Coriolis and magnetic forces similarities)

Edited by Aethelwulf, 11 October 2012 - 11:12 PM.

### #2 Aethelwulf

Aethelwulf

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• 936 posts

Posted 11 October 2012 - 07:00 PM

Darn it.... it uses math tex. This will take some time to fix... sorry about that.

### #3 CraigD

CraigD

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Posted 11 October 2012 - 07:11 PM

Darn it.... it uses math tex. This will take some time to fix... sorry about that.

We've got a LaTeX:math renderer associated with the math, imath, and I think, still, latex tags, so you should be able to find/replace fix this in a jiffy. Just change
$$\hbar c = GM^2$$
etc. to
$\hbar c = GM^2$
etc. and it should render correctly as
$\hbar c = GM^2$
etc.

### #4 Aethelwulf

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Posted 11 October 2012 - 07:36 PM

Sorry about that... the dizzy latex is fixed.

### #5 Aethelwulf

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Posted 11 October 2012 - 07:37 PM

We've got a LaTeX:math renderer associated with the math, imath, and I think, still, latex tags, so you should be able to find/replace fix this in a jiffy. Just change
$$\hbar c = GM^2$$
etc. to
$\hbar c = GM^2$
etc. and it should render correctly as
$\hbar c = GM^2$
etc.

Thank you, I didn't realize there where so many

### #6 Aethelwulf

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Posted 11 October 2012 - 07:43 PM

It has also came to my attention, that there is of course a consistency with the relationship

$E_gr_s = GM^2$

that I hadn't realized until writing about how Einstein speculated the inertia of a body (the gravitational charge) depended on energy. The uncertainty spoke about between the radius of a system and it's mass then depend on each other as such

$E_g \int r_s(M) = GM^2(r_s)$ *

which means... and this is going to sound completely absurd at first, but that the Planck Mass is truly not constant, when in light of the fact that uncertainty principle does many strange things like making such a condition of this mass vary depending on the measurement made on the radius. After all, you cannot be completely certain about a Planck Mass if it is highly uncertain in face of the radius of the particle you measure.

Also, keep in mind the limit equation

$\lim_{r_s \rightarrow 0} E_gr_s = GM^2 = 0$

The equation (*) is simply a generic integral form. Let's decide to use the limit of the integral in such a way similar to the calculus in our previous equation

$E_g \int_{r_s}^{0} r_s(M) = GM^2(r_s)$

The dependency relationship on $r_s$ is that it remains non-zero for the mass to exist. This says a great deal about massless particles in general, because massless particles do not have a Schwarzschild radius. And, so one can only conclude again, they cannot have a rest energy or on the right hand side, a gravitational charge (mass).

The uncertainty principle, if there is one between the radius and the mass of a Planck Particle, must somehow mean that the mass can indeed vary when a measurement is performed on it's complimentary observable (radius). But how this can be true, is a paradox, because we all know that the Planck Mass is in fact a constant... but is this always true in nature... can some constants... not be so constant?

There is a lot of evidence to suggest other constants in nature have changed for instance. When I find time, I will find references to show you how subtle changes for instance in the fine structure constant may have occurred, or the dynamics which makes the fine structure constant.

ref: http://physicsworld....-not-a-constant

Edited by Aethelwulf, 11 October 2012 - 11:22 PM.

### #7 Aethelwulf

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Posted 11 October 2012 - 07:50 PM

Fixed two equations which did not come out right as well.

Done job.

### #8 Aethelwulf

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Posted 11 October 2012 - 09:51 PM

I have just noticed that Motz has made a mistake in his paper

I followed some of his work, here:

$R^2 = (\frac{\hbar}{Mc})^2$

or simply from the original relationship

$\hbar = RMc$

as he mentions. He notes that one may look at this as an uncertainty relationship, between the rest momentum $Mc$ and the radius of the particle... Or put better, the space-interval it occupies. He determines that this must mean that the radius can be represented by a dynamical operator, that does not commute with the mass term $M$. Therefore the equation he proposed was

$\frac{R}{c}Mc = \hbar$

''

Well, the thing is, $\frac{R}{c}Mc = \hbar$ does not have the correct dimensions so what he says about the rest momentum cannot be true, however, if he said

$\frac{R}{c}Mc^2 = \hbar$

Then yes, the dimensions are correct now and actually fits perfectly into the modified second order Dirac Operator I presented with the invariant time operator... oh well.. a genius can't be right all the time ya know So what I will do is rewrite his equation properly without the talk of rest momentum, and rather a rest energy.

Notice the part

$E^2\frac{R^2}{c^2}$

$D \psi(x)= E^2\frac{R^2}{c^2} (\frac{12GM}{c^2})^{-2}\psi(x)$

To justify what I am saying.

Edited by Aethelwulf, 11 October 2012 - 09:59 PM.

### #9 Aethelwulf

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Posted 11 October 2012 - 11:33 PM

It has also came to my attention, that there is of course a consistency with the relationship

$E_gr_s = GM^2$

that I hadn't realized until writing about how Einstein speculated the inertia of a body (the gravitational charge) depended on energy. The uncertainty spoke about between the radius of a system and it's mass then depend on each other as such

$E_g \int r_s(M) = GM^2(r_s)$ *

which means... and this is going to sound completely absurd at first, but that the Planck Mass is truly not constant, when in light of the fact that uncertainty principle does many strange things like making such a condition of this mass vary depending on the measurement made on the radius. After all, you cannot be completely certain about a Planck Mass if it is highly uncertain in face of the radius of the particle you measure.

Also, keep in mind the limit equation

$\lim_{r_s \rightarrow 0} E_gr_s = GM^2 = 0$

The equation (*) is simply a generic integral form. Let's decide to use the limit of the integral in such a way similar to the calculus in our previous equation

$E_g \int_{r_s}^{0} r_s(M) = GM^2(r_s)$

The dependency relationship on $r_s$ is that it remains non-zero for the mass to exist. This says a great deal about massless particles in general, because massless particles do not have a Schwarzschild radius. And, so one can only conclude again, they cannot have a rest energy or on the right hand side, a gravitational charge (mass).

The uncertainty principle, if there is one between the radius and the mass of a Planck Particle, must somehow mean that the mass can indeed vary when a measurement is performed on it's complimentary observable (radius). But how this can be true, is a paradox, because we all know that the Planck Mass is in fact a constant... but is this always true in nature... can some constants... not be so constant?

There is a lot of evidence to suggest other constants in nature have changed for instance. When I find time, I will find references to show you how subtle changes for instance in the fine structure constant may have occurred, or the dynamics which makes the fine structure constant.

ref: http://physicsworld....-not-a-constant

Well, so let's ask the most obvious question... if mass is allowed to vary (the Planck Mass that is) what is varying fundamentally which defines it... is it $\hbar$, $c$... the gravitational constant perhaps $G$, since the Planck Mass is

$M = \sqrt{\frac{\hbar c}{G}}$

Well... I can be sure that the speed of light is not varying... that is always constant, one of the first principles you learn in relativity... $\hbar$ is the angular momentum... or the quantum action as I prefer calling it. It might vary... but unlikely... But what do we know about quantifying the gravitational constant at the fundamental quantized level?

That's a difficult question, because there is no leading or rather... overall consensus on how gravity does act at the appropriate Planck Scales. There is certainly no theory in quantum mechanics which can deal with it... sufficiently.

Edited by Aethelwulf, 11 October 2012 - 11:36 PM.

### #10 Aethelwulf

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Posted 11 October 2012 - 11:45 PM

The fusion of quantum theory, with that of the electromagnetic theory and that of gravitation requires what is called $(G,c,\hbar)$-physics. A good read is this

http://en.wikipedia....iki/CGh_physics

In fact, this very line of work is heavily involved in thermodynamics of black holes.

Edited by Aethelwulf, 11 October 2012 - 11:46 PM.

### #11 Aethelwulf

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Posted 12 October 2012 - 12:37 AM

Here is a ''paper'' which explains how $G^{-1}$ as a scalar field will allow a varying gravitational constant.

http://www.einstein-...s/scalar-tensor

Edited by Aethelwulf, 12 October 2012 - 12:38 AM.

### #12 Aethelwulf

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Posted 12 October 2012 - 01:12 AM

Any way, later on today I am going to give these previous questions a rest and start working on how matrices can explain the gravitational charge obeying Dirac Matrices. Good night.

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Posted 13 October 2012 - 01:29 PM

I lie... (changed my mind). I am not going to talk about the matrices yet because I found a new different form based on an uncertainty relationship between the temperature and time with the second order Dirac Operator.

The two equations of interest are

$D \psi(x)= E^2\frac{R^2}{c^2} (\frac{12GM}{c^2})^{-2}\psi(x)$

$\Delta \frac{GM^{2}_{p}}{\ell_p} \Delta t_p \propto k\Delta T(t_p) \frac{R}{c}$

Keep in mind that $kT$ has units of energy, this means we can further rewrite the Dirac Operator in a new form

$D \psi(x)= k^2 \Delta T_{p}^{2}(t_p)\frac{R^2}{c^2} (\frac{12GM}{c^2})^{-2}\psi(x)$

But because this part

$k^2 \Delta T_{p}^{2}(t_p)\frac{R^2}{c^2}$

is proportional to

$\Delta \frac{GM^{2}_{p}}{\ell_p} \Delta t_p$

This means we can restate the Dirac Operator as

$D \psi(x) = \Delta \frac{G^2M^{4}_{p}}{\ell_{p}^{2}} \Delta t_{p}^{2} (\frac{12GM}{c^2})^{-2}\psi(x)$

The invariant time operator no longer exists in this equation unless we see it as a ''form'' of what is called the Planck Time, which should be expected anyway since we are dealing with a Planck Scale system.

Edited by Aethelwulf, 13 October 2012 - 01:30 PM.

### #14 Aethelwulf

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Posted 13 October 2012 - 06:28 PM

The full interpretation of

$D \psi(x) = \Delta \frac{G^2M^{4}_{p}}{\ell_{p}^{2}} \Delta t_{p}^{2} (\frac{12GM}{c^2})^{-2}\psi(x)$

Is that, the change of the energy of a Planck Particle, in product of the time it takes to give up this energy is itself a coefficient of the Schwarzschild radius. The operator itself then, describes how such a particle can give up an energy in a temperature provided by kT.

Edited by Aethelwulf, 13 October 2012 - 06:29 PM.

### #15 Aethelwulf

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Posted 13 October 2012 - 06:59 PM

And now to the matrix forms.

As we can see here, we have written the canonical relativistic momentum form of $i\hbar \gamma^0$ as a matrix involving the gravitational charge by multiplying through by c.

$i \hbar c \gamma^0 = \begin{pmatrix} iGM^2 & 0 & & 0 \\0 & iGM^2 & 0 & 0 \\0 & 0 & -iGM^2 & 0 \\0 & 0 & 0 & -iGM^2 \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}$

$(\gamma^0)^2$ just punches out the unitary matrix, basically ignorable then, and squaring we remove the imaginary parts

$-\hbar^2 c^2 \mathbb{I} = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \cdot \gamma^0$

This matrix can be insightful because we now know that we have ended up with the the signs mixed with no imaginary coefficients. But of course, instead of writing it this way, we could have been much quicker by simply squaring everything

$-\hbar^2 c^2 = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2$

The matrix for $\gamma^0$ can be written this time

$-\hbar^2 c^2 \begin{pmatrix} 0 & \sigma^1 \\ \sigma^2 & 0 \end{pmatrix} = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}$

and since $\gamma^0 = \beta$ (the standard beta matrix) then $\beta a^k = \gamma^k$ thus

$-\hbar^2 c^2 \gamma^k = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}a^k$

$\gamma^k$ can now be written in it's sub-matrix form

$-\hbar^2 c^2 \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix} = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}a^k$

So calculating the very last matrices can be quite easy in fact and perhaps, quite boring

$-\hbar^2 c^2 \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix} = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}a^k$

Let's work with

$\begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}a^k$

and now write the matrix case for $a^k$ which is

$a^k = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix}$

so calculating it we get

$\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \end{pmatrix}$

To solve the right hand side completely, the final matrix form will look like

$-\hbar^2 c^2 \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -GM^2 & 0 \\0 & 0 & 0 & GM^2 \\0 & -GM^2 & 0 & 0 \\-GM^2 & 0 & 0 & 0 \end{pmatrix}^2$

Now this is just one result, by involving pauli matrices $\sigma_1$ and $\sigma_3$. We should solve the rest of matrix possibilities - this will take up a lot of space, so I will keep the calculations at a minimum. Before I even undertake that task, I want to show the equation above in a more simplified version:

$\begin{pmatrix} 0_2 & -\hbar^2c^2 \cdot \sigma^1 \\ \hbar^2 c^2 \cdot \sigma^3 & 0_2 \end{pmatrix} = \begin{pmatrix} 0 & 0 & -GM^2 & 0 \\0 & 0 & 0 & GM^2 \\0 & -GM^2 & 0 & 0 \\-GM^2 & 0 & 0 & 0 \end{pmatrix}^2$

where the full form of

$\begin{pmatrix} 0_2 & -\hbar^2c^2 \cdot \sigma^1 \\ \hbar^2 c^2 \cdot \sigma^3 & 0_2 \end{pmatrix}$

is

$\begin{pmatrix} 0 & 0 & 0 & -\hbar^2 c^2 \\0 & 0 & -\hbar^2 c^2 & 0 \\-\hbar^2 c^2 & 0 & 0 & 0 \\0 & \hbar^2c^2 & 0 & 0 \end{pmatrix}$

An interesting symmetry revealed itself in these matrix equations.

$\begin{pmatrix} 0 & 0 & 0 & -\hbar^2 c^2 \\0 & 0 & -\hbar^2 c^2 & 0 \\-\hbar^2 c^2 & 0 & 0 & 0 \\0 & \hbar^2c^2 & 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 0 & -GM^2 & 0 \\0 & 0 & 0 & GM^2 \\0 & -GM^2 & 0 & 0 \\-GM^2 & 0 & 0 & 0 \end{pmatrix}^2$

The right hand side is the complete mirror symmetry, only backwards as though one was to place a mirror to a sentence and find it one way reflected on the mirror and the other way presented on paper.

This symmetry is no accident. If we reduced the entries to unity (for the simplicity of investigating this symmetry further) we find that if you multiply the right hand side with the left hand side of this matrix equation) the symmetry comes back with real numbers

$\begin{pmatrix} 0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}$

What I noticed it is a submatrix with entries

$\begin{pmatrix} \sigma^1 & 0_2 \\ 0_2 & \mathbf{1}_2 \end{pmatrix}$

### #16 Aethelwulf

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Posted 13 October 2012 - 07:05 PM

I don't believe I have ever seen the matrix in such a form as this:

$\begin{pmatrix} \sigma^1 & 0_2 \\ 0_2 & \mathbf{1}_2 \end{pmatrix}$

I am going to be audacious enough to say that this will be called $\gamma^{1_{2}}$. There is a reason for this, because one can find this matrix through the use of the standard gamma matrices and this matrix by the formula

$\gamma^0\ \gamma^1 M = \gamma^{1_{2}}$

where $M = \begin{pmatrix} 0 & 0 & 0 & 1 \\0 & 0 & 1 & 0 \\1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \end{pmatrix}$

My gamma matrix has been given the name $\gamma^{1_{2}} = \begin{pmatrix} \sigma^1 & 0_2 \\ 0_2 & \mathbf{1}_2 \end{pmatrix}$ to represent the entry of the submatrix $\sigma^1$ and it's other entry, $\mathbf{1}_2$.

So let us just very quickly summarize what we have achieved. We have found that by formulating the matrix for the gravitational charge and multiplying it against $\beta a^k$:

$-\hbar^2 c^2 \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix} = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}a^k$

The Dirac Algebra can describe the symmetries of the entries via

$\gamma^0 \gamma^1 \bar{\gamma}^{1_{2}} = \bar{\gamma^{1_{2}}}$

And to finish, chirality can be found from the formula

$\gamma^0 \gamma^1 \bar{\gamma}^{1_{2}} = \bar{\gamma^{1_{2}}}$

To get back

$\gamma^0 \gamma^1$ you simply have to do this

$\gamma^0 \gamma^1 = \bar{\gamma}^{1_{2}} \bar{\gamma^{1_{2}}}$

Since Chirality is

$i \gamma^0 \gamma^1 \gamma^2 \gamma^3$

then it can be deduced we can get Chirality from my new formula

$\gamma^5 = i\bar{\gamma}^{1_{2}} \bar{\gamma^{1_{2}}} \gamma^2 \gamma^3$

### #17 Aethelwulf

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Posted 13 October 2012 - 07:08 PM

The matrix $M$ is then just

$\begin{pmatrix} \mathbf{1}_2 & 0_2 \\ 0_2 & \sigma^1 \end{pmatrix}$

So it's diagonal entries are reversed to the the other new gamma matrix

$\begin{pmatrix} \sigma^1 & 0_2 \\ 0_2 & \mathbf{1}_2 \end{pmatrix}$

I wouldn't even normally call them a gamma matrix if it was for the fact their product came to the same product as $\gamma^0\gamma^1$. There are as I have been informed, an infinite amount of equivalent Dirac Matrices, but these are concerned with the gravitational charge.