## Planck Particles

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Simon Belmont
Posts: 38
Joined: Fri Sep 14, 2012 10:58 am

### Planck Particles

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 graviational 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.de/mirror/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.thefreedictiona...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 $e$.

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.

We may start with the quantum relationship

$\hbar = RMc$

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

$\hbar c = GM^2$

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.gravityresearchfoundation.or ... 1/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 $-2M\omegav$ (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 (1))

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

$e(v \times B) = M(\omega \times v)$

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.

Another way to write this equation is in terms of the Gauss' Law for gravitation. One may see that the radius of curvature can be obtain from the Gauss' law by dividing the speed of light squared on both sides

$\nabla \cdot \frac{g}{c^2} = -8\pi \rho S_c$ plugging this into the equation we get

$-\frac{2\omega}{\sqrt{G}} \times (\omega \times \nabla \cdot \frac{g}{c^2}) = B \times v$

Where $g$ is the gravitational field. One must keep in mind that the Gaussian curvature is an intrinsic property of space and of course, so is the fields we are dealing with in this formula. What maybe of further interest, is that earlier I said it is not generally suitable to describe Planck Particles gravitational charge by the conventional formula

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

However one can arrive at a very similar kind of equation for the gravitational charge using the Gauss' integral form by a small change, that being the mass has now been multiplied on both sides yielding

$\int \int_{\partial A} M(g \cdot dA) = -8 \pi GM^2$

Since the $M$ in this work remains constant, we can pull that outside of the integral kernel

$M \int \int_{\partial A} (g \cdot dA) = -8 \pi GM^2$

which gives the gravitational charge on the right hand side, the quantity I wished to keep invariant in the equations when describing these exotic particles. Another reason perhaps to favour this form over the the conventional form, is that the modified Gauss equation describes the area of the system, whilst the conventional equation calculates the charge contained on the 2D surface. That wouldn't be beneficial for this work since the gravitational charge and electric charge occupy inside the sphere, or encompass inside the shell of radius that we have adopted to model the work with.

(1) http://www.gravityresearchfoundation.or ... 1/motz.pdf

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

(3) http://arxiv.org/ftp/arxiv/papers/1109/1109.3624.pdf (Coriolis and magnetic forces similarities)
Last edited by Simon Belmont on Wed Sep 19, 2012 8:55 am, edited 4 times in total.

Simon Belmont
Posts: 38
Joined: Fri Sep 14, 2012 10:58 am

### Re: Planck Particles

I wish to note that $E_g$ which I have identified as a gravitational energy, still has units of energy. Gravitational energy in the work is taken to be synonymous with inertial energy, which is the energy which is associated to the quantity of gravitational charge (mass of the system).

Simon Belmont
Posts: 38
Joined: Fri Sep 14, 2012 10:58 am

### Re: Planck Particles

My latest thoughts on the equations in the previous post (OP) is now trying to find relationships between the gravitational charge and the idea of spinning spheres, naturally since we have been concerned with such systems. The reader is will either find these new equations interesting, or not very interesting.

We begin with the magnetic moment definition related to spin and the g-factor. I actually arrive at around three different equations describing the spin relationship with the gravitational charge.

$\mu = \frac{g_s e}{2m}S$

Rearranging and multiplying $\sqrt{G}$ we obtain

$2 \sqrt{G}M \mu = \sqrt{G} g_s e S$

divide off $2 \mu$ on both sides gives

$\sqrt{2 \mu}$ we obtain

$\sqrt{G}M = \frac{\sqrt{G}g_s e S}{2\mu}$

Since in Dirac's theory, (and we are dealing with spin 1/2 particles when concerning Planck Particles), we find that $g_s = 2$ in quantum theory strictly speaking and because of this, the equation simplifies to

$\sqrt{G}M = \sqrt{\hbar c} = \frac{e S}{\mu}\sqrt{G}$

which is our first new equation relating spin and charge as being proportional to the gravitational charge. For the Pauli Spin matrices, you would have three components related to the cartesian coordinates

$\sqrt{G} \frac{e}{\mu} \frac{\hbar}{2} \vec{\sigma}_x = \sqrt{G}M$

$\sqrt{G} \frac{e}{\mu} \frac{\hbar}{2} \vec{\sigma}_y = \sqrt{G}M$

$\sqrt{G} \frac{e}{\mu} \frac{\hbar}{2} \vec{\sigma}_z = \sqrt{G}M$

One last relationship, is based similarly to our beginning derivation. We begin with

$2M \mu = g_s e S$

multiply $\sqrt{G}$ on both sides, motivation behind this is to obtain the gravitational charge

$2\sqrt{G}M \mu = \sqrt{G}g_s e S$

now we realize that

$E = -\mu \cdot B$

Which is the Larmor energy. Because of this relationship, we choose to take the dot product of the magnetic moment with the magnetic field (which must be kept in mind has three components).

$2 \sqrt{G}M(-\mu \cdot B) = (\nabla \times A)\sqrt{G} g_s e S$

dividing off $2\sqrt{G}M$ both sides yields our final relationship for this post

$E = -\mu \cdot B = (\nabla \times A)\frac{eS}{M}$

Simon Belmont
Posts: 38
Joined: Fri Sep 14, 2012 10:58 am

### Re: Planck Particles

So based on a previous equation in the past post, we ended up with

$\mathcal{H} = -\mu \cdot B = \frac{eB}{M} S = (\nabla \times A)\frac{e}{M}S$

Since of course we are dealing with Planck Particles, the ratio

$\frac{e}{M}$ is in fact the ratio of the elementary charge to the Planck Mass. Calculating the value to this ratio we obtain:

$\frac{e}{M_p} \propto 1.861 721 726 984 943 \times 10^{-10} C/kg$

Simon Belmont
Posts: 38
Joined: Fri Sep 14, 2012 10:58 am

### Re: Planck Particles

What is the justification for a mass to charge, or charge to mass ratio?

The answer lye's primarily in the Lorentz Force law. Quickly reciting the equation,

$F = e(v \times B)$

If one take the force as $Ma$ which is the definition of a force, then by a simple rearrangement of the elementary charge one can find a mass to charge ratio as

$\frac{M}{e}a = (v \times B)$

Where $a$ is the acceleration naturally, $E$ is the respective electric field, $v$ is the linear velocity and $B$ of course is the magnetic field. The Hamilatonian equation I derived was

$\mathcal{H} = -\mu \cdot B = \frac{eB}{M} S = (\nabla \times A)\frac{e}{M}S$

where we have an inverse charge to mass ratio.

$\Delta E = -\mu \cdot B$

Whilst this is generally the energy of the system, the sign (-) has strong correlations to spin dynamics which shows up in the term $S$ due to Planck Particle spin flip energy difference. This means that it can be applied to a free Planck Particle with a spin projection in the so-called, magnetic flux density. By doing do, we are attempting to describe gravitationally influential particles in some magnetic field unification, which is generally the same approach we have been trying to describe when united gravimagnetic fields for spinning micro black holes.

Could an inverse charge to mass, mass to charge relationship have anything to do with the energy difference? One might think so, but it must be noted imperatively that this is only a hand-waving speculation.

This newest post gives rise to some new relationships based on the original paper concerning the gravimagnetic relationships for spinning spheres. The corriolis force (the gravimagnetic field)'s acceleration is given as

$a = (\omega \times v)$

Plugging this into the equation

$\frac{M}{e}a = (v \times B)$

yields

$\frac{M}{e}(\omega \times v) = (v \times B)$

Since we have ascertained that the Motz definition of the Gravimagnetic field is

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

then we can further demonstrate this is the same as saying

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

Multiplying $\sqrt{G}$ on both sides gives us a relationship with the fine structure constant by squaring the whole set of equalities

$\frac{GM^2}{e^2}(\omega \times v)^2 = \frac{GM^2}{\hbar c} a^2$

Where $\frac{GM^2}{e^2}$ is recognized as the fine structure constant.

Similarities to the Zeeman Effect and the Magnetic Moment Definition Proportional to Spin and my Hamiltonian Equation

The Zeeman Effect is an energy term given as

$\Delta E = \frac{e \hbar B}{2M}$

The Magnetic Moment is defined as

$\frac{g_s B}{2M} S = \mu$

The Hamiltonian variant which uses some of these concepts is given again as

$\mathcal{H} = -\mu \cdot B = \frac{eB}{M} S = (\nabla \times A)\frac{e}{M}S$

Notice in the equation the expression

$\frac{eB}{M} S$

This was obtained in the understanding through relativity that the correct g-factor for particles at (near) rest is given as $g_s$ is actually just a factor of $2$ which cancelled the factor of $2$ in the denominator, which is how we ended up at the expression above. Perhaps, interesting or not, we can take the final equation in the first post in a more complicated form:

$-\frac{2\omega}{\sqrt{G}} \times (\omega \times 8 \pi \rho_0 (\frac{G}{c^2})) = \frac{M}{e}a$

This unites the idea that there is indeed a mass-to-charge ratio for a Planck Particle where the Gaussian curvature and the gravimagnetic field is concerned.

Simon Belmont
Posts: 38
Joined: Fri Sep 14, 2012 10:58 am

### Re: Planck Particles

Proving that the Schwarzschild Radius is itself Invariant for a Non-Zero Gravitational Charge (Mass) $\sqrt{G}M$ In a Different Outlook

In my original post, which can be found as the last post to all the other posts, I contended that the relationship

$E_g r_s = GM^2$

was a type of definition of what gravitational mass was. Whilst $E_g$ still had units of energy, I defined it specially as a ''gravitational energy.'' It was a new name I derived for a new type of concept which attempted to bridge the idea of what inertial mass is and what Motz had in mind when he considered inertial mass as a gravitational charge.

Since in my work, I defined the gravitational charge in a spin relationship given as

$\sqrt{G} \frac{e}{\mu} \frac{\hbar}{2} \vec{\sigma} = \sqrt{G}M$

we will now concentrate on the left hand side in a completely new way, but by assuming that the quantity $E_g r_s$ is not entirely complete, instead, should be seen as an integral over the energy in respect of the Schwarzschild radius $r_s$ and by setting it equal to the left hand side expression we just mentioned. This then takes the form of

$\int E_g\ dr_s = \sqrt{G} \frac{e}{\mu} \frac{\hbar}{2} \vec{\sigma}$

Before, I explained

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

How if the radius goes to zero, we cannot speak about a gravitational intrinsic mass. Now what is interesting from the approach made in the equation above is that if

$\int_{r_s}^{0} E_g\ dr_s = \sqrt{G} \frac{e}{\mu} \frac{\hbar}{2} \vec{\sigma}$

then this directly effects the right hand side by saying that

$r_s \rightarrow 0$

(which we established) then so does the elementary charge

$e \rightarrow 0$

via this equation

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

but if charge goes to zero, the consequence of this equation is that it cannot either describe $\frac{\hbar}{2} \vec{\sigma}$ since the coefficient of the angular momentum is the charge and if the charge is zero, then so is the spin which strictly describes spin 1/2 particles. All these quantities which are related to the gravitational charge implies there can be no inertial gravitational charge, or inertial mass, whatever one wishes to call mass. The implications can be seen in a more productive sense, because if the radius goes to zero, then we cannot talk about Planck Particles, or any type of particles described by spheres because the bare mass of such a particle would become infinite. There for, only massless radiation can be described as being pointlike systems, because they do not contain a bare mass. Renormalization procedures are often adopted to solve this incongruity, by turning the bare mass of a particle negative. However, if we treat particles with mass as a special condition synonymous with the Schwarzschild radius, then we can say it does not have an infinite bare mass without resorting to altering the bare mass as a negative quantity. Photons do not have such a radius and therefore should not be subject to infinite energies.

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### Re: Planck Particles

Two Objections by a PhD

''1. At the Planck scale talking about the electric and magnetic fields is wrong, they no longer exist in that form by to electroweak unification (an observed experimental fact).
2. There is no Planck scale model for any of the forces so none of the coefficients you're using have been justified to be valid at such energy scales''.

Just because a man may have a PhD does not make specific objections correct. My retorts to these objections are as follows. The first and second object as based on similar fallacies which have not been considered.

Refutation 1

Actually, at the Planck Scale talking about unification of the forces must be implemented. The reasons are simple. First of all, you cannot call it a unification of weak or electric if they are supposed to be unified into a single force. Electroweak unification does not mean that these forces are not present, but rather they are unified into the Langrangian - unification does not mean an entirely new force per se, it just means that the unification has mixed field terms. If one actually studies the electroweak Langrangian, one will find that the Langrangian composes of not only electric field currents but also the magnetic field currents. This means that the Langrangian does indeed have the electromagnetic fields present within the theory, they are not devoid as the first objection seems to indicate. In a more fuller form, the electroweak Langrangian is actually made up of five other components as well, the Langrangian due to Gauge particles,

$W^{i}_{\mu \nu} = \partial_{\mu} W^{i}_{\nu} - \partial_{\nu}W^{i}_{\nu}ig \epsilon^{ijk}W^{i}_{\mu}W_{\nu}^{k}$

also made up of

$B_{\mu \nu} = \partial_{\mu}B_{\nu} -\partial_{\nu}B_{\mu}$

It is also made up of $L_{matter}$, $L_{Higgs}$, $V{\phi}$ and $L_{Yukawa}$

Just so that there are no misunderstandings, $W_{\mu \nu}$ and $B_{\mu \nu}$ are field strengths akin to electromagnetism $F_{\mu \nu}$. So certainly the math and even the Langrangian spit's out mixed terms.

So how one can say they play no role in the electroweak unification for Planck Scales, is beyond me. Clearly the energy of the Langrangian has mixed terms of the forces required to describe this work. It is like electromagnetism. Just because electric and magnetic fields have been unified, does not mean they are not present. A moving charge for instance still experiences a magnetic field in a current, a topic I will be touching on in my next post.

Refutation 2

Physicists have a fairly good understanding of fundamental interactions on the quantum level, however gravity is problematic, and cannot be integrated with quantum mechanics at very high energies using the usual framework of quantum field theory. And whilst there is no Planck Scale model for such interactions at this time, is not meant to be there will not be in the future, only the future will tell if such particles like Planck Particles do experience gravimagnetic forces. It is premature to think that the work I have been doing is invalid when considering the unification of gravity with the other forces at this level is so aloof but time will tell and I would wager that most scientists would consider it possible.
Last edited by Simon Belmont on Sat Sep 22, 2012 6:35 am, edited 3 times in total.

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### Re: Planck Particles

My next mathematical work will be to derive the Larmour electromagnetic inertia term involved in Motz's modification to describe the brightness of quasars. Whilst my work will not specifically involve the theory of quasars, it will bring into some interesting proposals that are in contrast with analogies involved between elementary charges that radiate while being accelerated, to that being proportional to inertial charges (particles with a gravitational charge) also giving rise to such phenomenon. This phenomenon is most recognized by the name as the Larmour equation. Motz' paper can be found here:

http://gravityresearchfoundation.org/pd ... 8/motz.pdf

My work will be presented in the next post, this was just to provide a quick summery.

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### Re: Planck Particles

A Gravimagnetic Charge Radiating

Now, the Larmor equation is

$P = \frac{2}{3} \frac{e^2}{c^3}a^2$

This equation has dimensions of power radiation by an electron of charge when accelerated.

Schwinger presented a relativistic version of this equation as (as noted in the paper by Motz as)

$\mathbf{P} = \frac{2}{3} \frac{e^2}{M^2c^3} (\frac{d P}{ds})$

The Schwarzschild shell expression for $ds$ is

$(1 - \frac{GM}{c^2 r}) dt^2$

so we may plug this into the power radiated by a charge around a spherical body of mass as

$\mathbf{P} = \frac{2}{3} \frac{e^2}{c^3} \frac{1}{(1 - \frac{GM}{c^2 r})}(\frac{d P}{ds})$

which further becomes

$\mathbf{P} = \frac{2}{3} \frac{e^2}{c^3} \frac{a^2}{(1 - \frac{GM}{c^2 r})} dt^2$

The Schwarzschild shell expression can be written in terms of a gravitational charge over the integral of energy in respect of the radius $r_s$

$\mathbf{P} = \frac{2}{3} \frac{e^2}{c^3} \frac{a^2}{(1 - \frac{GM^2}{\int E\ dr_s})} dt^2$

Concerning the shell theory of the Planck Particle, and unusual problem does begin to present itself. The elementary charge as we have gathered a number of times now, comes in a relationship with the form of

$e = \sqrt{4 \pi \epsilon GM^2}$ in SI units. But what I did was something, which I have never seen in the archives of physics, is by treating this shell metric with an intrinsic gravitational charge term

$ds^2 = (1 - \frac{GM^2}{\int E\ dr_s}) dt^2$

which of course $(1 - \frac{GM^2}{\int E\ dr_s})$ is a dimensionless quantity. Just as a side note, only dimensionless quantities have real physical meaning in physics). Anyway, the Schwarzschild solution is prevalent for stagnant, non-spinning particles, or even macroscopic celestial objects. This means that they do not actually have a charge $e = 0$ and if the elementary charge is zero for a non-spinning Planck Particle, then that would imply that the mass (the gravitational charge $\sqrt{G}M$) must also by default sum to the fat big goose egg, nada, absolute zero.

But this is a real problem and in the face of mathematical physics describing these objects, since after all, black holes have mass so how can spin-zero Planck Particles exist in nature? The way out of this, I give a proposal, one of four.

1) That black hole Planck particles are never found in spin zero states.
2) Or, that the Schwarzschild shell are inadequate to describe these particles.
3) Or both the above, since 1) relies on 2) and 2) is a consequence of our first option.
4) Or ( and now this is pure speculation) but might it be possible to have a zero elementary charge, but still have mass through the presence of magnetic moments? This is speculated because the Neutrino has no electric charge, but current theory suggests it has a very small mass and also is a fermion particle (spin 1/2 dynamics).
Last edited by Simon Belmont on Thu Sep 20, 2012 11:31 pm, edited 5 times in total.

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### Re: Planck Particles

The interpretation of Two Previous Equations

$\mathbf{P} = \frac{2}{3} \frac{e^2}{c^3} a^2$

This equation is the Electromagnetic inertia, but the term electromagnetic inertia comes at a price. This isn't just about accelerating a particle like an electron, it also involves all kinds of electrically-charged particles. Planck Particles are themselves subject to Unruh-Hawking radiation, the smaller they are, the faster they give up their mass. There could be a theoretical possibility that you may feed such a particle mass is if it was immersed in a dense fluid so that it did not simply give up it's radiation in the expected Planck Time. The justification for doing this, is to see it's behaviour when being accelerated in a fluid. It may give off a Larmor radiation just as we call it electromagnetic inertia, it is itself a phenomenon related to inertial mass via first principles. This is the so-called electromagnetic mass, the contention that electromagnetic energy contributes to some proportion of the inertial mass of a particle. The power generated by such a particle can be seen in terms of the mass of the object which is embedded in the Schwarzshild shell, which is itself an inertial energy, or as I have called it, a gravitational energy which is the same energy contribution idea to the gravitational or inertial mass. It is this important modification $(1 - \frac{GM^2}{\int E\ dr_s})$ I made to the metric expression which makes this interpretation possible, where I stress

$\int_{r_s} E\ dr_s = GM^2$

So it appears in the denominator as

$\mathbf{P} = \frac{2}{3} \frac{e^2}{c^3} \frac{a^2}{(1 - \frac{GM^2}{\int E\ dr_s})} dt^2$

So this is the radiation emitted by a charge round a large spherical distribution of gravitational charge.
Last edited by Simon Belmont on Wed Sep 26, 2012 6:34 am, edited 7 times in total.

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### Re: Planck Particles

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, Compton.

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.
Last edited by Simon Belmont on Wed Oct 10, 2012 1:51 pm, edited 3 times in total.

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### Re: Planck Particles

A Universe Once Flooded with Planck Particles

A gravitational charge does not need to be necessarily account for one charge. In my next three or so posts, I will now try and describe what is called a mass flow where we may decide to account for a universe which was once flooded in these primordial particles.

The mass flow equation I will use to describe this can be written in two ways:

$\dot{m}_i \psi = (\frac{\partial \mathcal{L}}{\partial \dot{q}_i}) d_i \nabla^2 \psi$

Where $d_i$ is our displacement of our particles and $\frac{\partial \mathcal{L}}{\partial \dot{q}_i}$ is our classical canonical momentum part where $q_i$ sums over all $i-$th particle velocities. This equation then has dimensions of a mass flow rate [6]. The time dependancy arises on the left hand side of the equation, but the right handside has a generalized velocity. The other way to write this

$\dot{m}_i \psi(x) = (i\hbar \gamma^0) d_i \nabla^2 \psi(x)$

Where the $\gamma^0$ is the time-dirac gamma matrix. In fact, the mass flow equation (the equation describing a many-body system of gravitationally-charged particles can be written a number of ways, this next way is purely quantum mechanical

$\dot{m}_i\psi = i\hbar \nabla \psi$

To calculate the overall the gravitational charge of all the constituents of this matter field of Planck Particle's is by recognizing we must take into consideration that we are carefully measuring the ith-sum of all the masses. Naturally, we do hot have the squared G-factor to give our true quantum definition of the gravitational charge, but this can be done quite easily.

In terms of a statistical analysis, you can view the flow of mass as a net flow rate as

$\sum^{k}_{k=1} \dot{M}_i \hat{S}_k = \nu_i$

Which is purely classical (Newtonian). In the next post, I will show you how to derive my mass-flow equation. This however does not need to describe Planck Particles all the time, we may choose any number of inertial force fields composed of particles with mass.

The gravitational mass flow equation can also be understood in terms of the function which defines entropy in a very neat set of equivalences

$\sum^{k}_{k=1} \dot{M}_i \hat{S}_k \psi(x) = \sum_{k=1}^{k} (i\hbar \gamma^0)_k d_i \nabla^2 \psi(x) = \dot{M}_k(-k_B} \sum_i P_i\ ln P_i) \psi(x)$

So the very statistical nature provides a method of us to calculate my mass flow equation in a number of ways, and the Plancks Constant provides a way or possibly produces a gap to make the equation semi-classical.
Last edited by Simon Belmont on Wed Sep 26, 2012 4:12 am, edited 9 times in total.

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### Re: Planck Particles

Wordlines, Gravitational Mass Flow Equations and Defining a New Quantum Field

When we think of a ''bunch'' of particles flowing (in quite possible many directions even at the one time) a physicist will usually consider calculating their worldlines.

First of all it seems best to explain why the term $\frac{\partial \mathcal{L}}{\partial \dot{q}}$ is important since this Canonical momentum appears in my mass flow equation and is also pivotal when describing world lines.

Consider a simple spacetime interval as:

$d\tau^2 = dt^2 - d\vec{x}^2$

Where we have set $c=1$ in this case. You actually calculate the length of a worldline by taking into consideration the integral

$L(W) = \int_W d\tau$

You can, it was shown to me a while ago now, that worldines can be written in terms of time by the chain rule. Doing so, you can rewrite the time derivatives as dots on your variables and can end up with

$L(W) = \int_{t_0}^{t_1} \sqrt{1 - ||\dot{x}||^2}\ dt$

From here, you would calculate the Langrangian by simply multiplying mass into the equation, so we would have

$\mathcal{L} = -M\sqrt{1 - ||\dot{x}||^2}$

Now in my equation, we have been using the generalized velocity, and can be freely exchanged now to make the above equation into

$\mathcal{L}(\dot{q}\dot{q}) = -M\sqrt{1 - \dot{q}\dot{q}}$

Now, the canonical momentum part in my equation can be written as

$\frac{\partial \mathcal{L}}{\partial \dot{q}} = \frac{M\dot{q}}{\sqrt{1 - \dot{q}\dot{q}}}$

This is relativistic and is incomporated as one can see, into the idea of worldlines. Now, in my equation, I decided to multiply the momentum with distance. Of course, this was just the quantum action $\hbar$, but ignoring that fact for now, we wish to calculate the distance really as a displacement of all the particles in the universe $d_i$ using Barbour's approach. Doing so, will require an integral.

Taking the integral of the equation, which ''cuts up'' or ''slices'' a worldline for a particle, then the distance will be small $\delta d$ for a particle which is the way elluded to in the OP for how to calculate displacement of particles instead of distance exactly.

Remembering that

$\frac{\partial \mathcal{L}}{\partial \dot{q}} = \frac{M\dot{q}}{\sqrt{1 - \dot{q}\dot{q}}}$

then my equation

$\frac{\partial \mathcal{L}}{\partial \dot{q}_i} d_i \nabla^2\psi = \dot{m}\psi$

Can actually be rewritten (including the integral this time) as

$\int \dot{m}\psi\ dt = \int (\frac{M\dot{q}}{\sqrt{1 - \dot{q}\dot{q}}}) \delta d_i \nabla^2\psi\ dt$

So as you can see, the integral not only ''cuts'' up the worldline of a particle appropriately, but making the interval short enough will ensure that your distance is really just a very small displacement on a system of particles.
Last edited by Simon Belmont on Sat Sep 22, 2012 8:14 am, edited 3 times in total.

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### Re: Planck Particles

If the Most Primordial Particle was a Planck Particle

This little excerpt slightly controversial for a number of reasons which we will not go into depth in this paper, but it does seem that the energy requirements at Big Bang where perfect to create a ''field'' of primordial Black Holes. Whilst large black holes are almost certainly permitted, and most will be close or have already given up their mass in the form of radiation, I wish to define a fifth kind of quantum field, the Planckion field, named after Planck's enigmatic theory of Planck particles.

To realize a new field, we must give it a symbol... and with much considerable thought, I have decided to call it $\Xi$.

It is always hard nowadays to find Greek Letters which haven't been used frequently, but we hardly ever come across this one. So the flow equation becomes

$\int_{ \Xi_i} \dot{\Xi}\psi dt = \sum_{\Xi = 1} (\frac{\X_i \dot{q}}{\sqrt{1 - \dot{q}\dot{q}}}) \Delta d_i \nabla^2\psi\ d( \Xi)$
Last edited by Simon Belmont on Wed Sep 26, 2012 4:10 am, edited 5 times in total.

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### Re: Planck Particles

How to Find the Rest Energy (of Planck Particles) in Conjecture of these Previous Works

How would one describe the energy then of a single particle in context of the previous equations? The energy of a single particle can be thought of as

$E_i\psi = c^2 \int \dot{m_p}_i \psi\ dt_p$

And perhaps even then, a simple Langrangian might have the form

$\mathcal{L} = (\frac{1}{2}\dot{r} \cdot \dot{r}\int \dot{m}\ dt - V)\psi$

What we have done for the Planck Equation, we rewrite the energy with a subscript $i$ to denote ''inertial energy'' - I could have also wrote it as $E_g$ but I have been frequently concerned whether this might confuse the reader, the integral has been made on all $i-th$ particles and the dummy variable also also been written in Planck Time. It did occur to me recently that we may even define the Planck Time as the shortest interval of an invariant time operator $\frac{R}{c}$, somewhat inspired by Motz from his paper on the second order Dirac Operator in which he finds the $R$ as the square root of the spacetime interval, defining the geometry of a particle which (has not only came into consideration in my work previously) but will continue to play a role when we take a more complicated look at the second order differential operator. Oh by the way, in concern of replacing the Planck Time for the invariant time operator, the equation would take the form:

$E_i\psi = c^2 \int \dot{m_p}_i \psi\ d(\frac{r_s}{c})$

We could even now decide to canonically quantize the equation

$i\hbar \frac{\partial}{\partial t_p} \psi = c^2 \int \dot{m_p}_i \psi\ d(\frac{r_s}{c})$

Which I think is quite a beautiful looking equation, but also knowing that the quantum action and any displacement must not only be in a Plank Time, but the displacement must equal the Planck length $\ell_p$ or be approximated to, depends on which take you wish to scrutinize the values.

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### Re: Planck Particles

A Gravitational Charge Operator from a Mass Flow Field Part I

$\dot{m}_i \psi = (\frac{\partial \mathcal{L}}{\partial \dot{q}_i}) d_i \nabla^2 \psi$

Take the integral of both sides of the equation in respects of the proper time

$\int \dot{m}_i \psi ds = \int (\frac{\partial \mathcal{L}}{\partial \dot{q}_i}) d_i \nabla^2 \psi ds$

Now we have

$\int \dot{m}_i \psi ds = m\psi$

Multiply the $\sqrt{G}$ we have

$\sqrt{G}\dot{m}_i \psi ds = \sqrt{G}M \psi$

Hit this with $\frac{\partial}{\partial t}$ and we ge

$- i\hbar c \frac{\partial}{\partial t} \psi = \sqrt{G}M \frac{\partial}{\partial t} \psi$

The right hand side is simply the energy operator $-i\hbar \frac{\partial}{\partial t} \psi$ multiplied by the speed of light. I believe, but not with a whole great deal of justification that

$\sqrt{G}M \frac{\partial}{\partial t} \psi$

Is the gravitational charge operator. You can also get this operator from the previous post in our last equation

$i\hbar \frac{\partial}{\partial t_p} \psi = c^2 \int \dot{m_p}_i \psi\ d(\frac{r_s}{c})$

Simply multiply the speed of light on both sides of the equation gives

$\sqrt{G}M \frac{\partial}{\partial t} \psi = c^2 \int \dot{m_p}_i \psi\ d r_s$
Last edited by Simon Belmont on Wed Sep 26, 2012 2:27 am, edited 10 times in total.

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### Re: Planck Particles

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 s 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 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 $x$ 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

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

..Part III coming up of the same subject posts.
Last edited by Simon Belmont on Fri Sep 28, 2012 3:42 am, edited 8 times in total.

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### Re: Planck Particles

Part III, the final part to the work concerning the the past 3 posts I will give you later today as it involves a lot if latex. However, a quick summery is that I will prove that in a special 2-particle universe, the gravitational equilibrium is conserved. I will demsonstrate this work on Hibert spaces, using direct tensor product in order to obtain a Hermitian Operator $h$ in which the particles can be described as a complete Orthornormal set. As this will be more of a mathematical exercise, the final investigation to this will be taking the Dirac Operator and explaining it in terms of how Motz explains that the $R$ the the square root of the spacetime interval. In my work, we will give this as $\sqrt{d^t^2 - d \vec{x}^2}$ and then see this as not only the geometry, but we must also conlude it is the source of the gravitational field. We will also investigate the Dirac Operator in terms of of the Weyl Limit and how further how $M = 0$ is not compatible.This will involve some difficult math but I will try and keep it as simple as possible, for my sake as well

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### Re: Planck Particles

...
Last edited by Simon Belmont on Mon Sep 24, 2012 8:21 pm, edited 2 times in total.

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### Re: Planck Particles

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.
Last edited by Simon Belmont on Wed Oct 10, 2012 1:50 pm, edited 3 times in total.

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### Re: Planck Particles

Hamiltonian constraints exist, ours exist in the form

$\hat{H}|\psi> = \hat{E}|\psi>$

where the constraint may play a part in my work later... what is a Hamltonian constraint?

$\pi_t + H = 0$

where $\pi_t$ is the canonically-conjugate momrntum time (ref.1)

(ref.1) - http://www.platonia.com/complex_numbers.pdf

later edit

''It can be seen as the generator of system evolution in time. It can also be seen as the conjugate momentum to time, meaning that if time is thought of a generalized coordinate, the momentum corresponding to that coordinate will be the Hamiltonian, just as the momentum corresponding to a Cartesian coordinate is the familiar linear momentum.''

source; everything2.com
Last edited by Simon Belmont on Mon Oct 01, 2012 2:28 pm, edited 2 times in total.

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### Re: Planck Particles

Simon Belmont wrote:A Gravitational Charge Operator from a Mass Flow Field Part II

$\sqrt{G}\dot{m}_i \psi ds = \sqrt{G}M \psi$

Hit this with $\frac{\partial}{\partial t}$ and we get

$- i\hbar c \frac{\partial}{\partial t} \psi = \sqrt{G}M \frac{\partial}{\partial t} \psi$

The right hand side is simply the energy operator $-i\hbar \frac{\partial}{\partial t} \psi$ multiplied by the speed of light. I believe, but not with a whole great deal of justification that

$\sqrt{G}M \frac{\partial}{\partial t} \psi$

Is the gravitational charge operator. The charge operator may actually be complexified since the energy operator has a $-i$ coefficient.

This equation as turns out

$- i\hbar c \frac{\partial}{\partial t} \psi = \sqrt{G}M \frac{\partial}{\partial t} \psi$

implies a different outlook on the Canonical gravitational charge operator as being non-conmplexified. Quantum gravity is believed to be non-complex. Reasons why is because quantized gravity yields to equations which are not complexified. This gravitational operator does not have complexification.

Keep in mind also that the operator can also show up in the mass flow equation

$\sqrt{G}\dot{m}_i \psi = \sqrt{G}(\frac{\partial \mathcal{L}}{\partial \dot{q}_i}) d_i \nabla^2 \psi$
Last edited by Simon Belmont on Fri Sep 28, 2012 3:41 am, edited 3 times in total.

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### Re: Planck Particles

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

$-\hbar^2 c^2 \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix} = \begin{pmatrix} -GM^2 & 0 & & 0 \\0 & -GM^2 & 0 & 0 \\0 & 0 & GM^2 & 0 \\0 & 0 & 0 & GM^2 \end{pmatrix}^2 \begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}a^k$
Last edited by Simon Belmont on Mon Oct 01, 2012 1:57 pm, edited 2 times in total.

Simon Belmont
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### Re: Planck Particles

Sorry got some of my signs mixed up in the matrices, fixed now...

Simon Belmont
Posts: 38
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### Re: Planck Particles

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

$\begin{pmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & 1 & 0 & 0 \\1 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\0 & 0 & 0 & -1 \\0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \end{pmatrix}$
Last edited by Simon Belmont on Mon Oct 01, 2012 1:44 pm, edited 1 time in total.

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### Re: Planck Particles

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

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

Simon Belmont
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### Re: Planck Particles

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

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

Again, stressing they are indeed unitary but also Hermitian.

Simon Belmont
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### Re: Planck Particles

Simon Belmont wrote: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

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

Again, stressing they are indeed unitary but also Hermitian.

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, $\mathbf{1}_2$.

Simon Belmont
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### Re: Planck Particles

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

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

Simon Belmont
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### Re: Planck Particles

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

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

Simon Belmont
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### Re: Planck Particles

I never actually said what $M$ was but judging its matrix I thought it was obviously. But just to write it out:

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

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 $\gamma^0\gamma^1$.

Simon Belmont
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### Re: Planck Particles

It turns out that my matrices will not satisfy the Dirac Algebra but it they are equivalent to a degree. It seems that the $1_2$ and $\sigma^1$ matrix entries which play the particle antiparticle roles would require a new interpretation.

Simon Belmont
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### Re: Planck Particles

Simon Belmont wrote:It turns out that my matrices will not satisfy the Dirac Algebra but it they are equivalent to a degree. It seems that the $1_2$ and $\sigma^1$ matrix entries which play the particle antiparticle roles would require a new interpretation.

Talking of equivalent matrices, there are of course, an infinite amount of them. I pose a question... if the trace of the matrix is zero, then the matrix determinant must always be $1$. Does this have any significance to what mass is?

Mathematics is so abstract it's hard to tell nowadays how to properly interpret an equation physically. The only satisfying conclusion is that matrices with non-zero traces must be something $\ne det|1|$. It would be these kind of matrices which describe massless particles. But this is a speculation and my mathematical capabilities... are limited at best when dealing with understanding how that can be implemented, but I am certainly trying to figure out if it can.

So without any more speculation on the matrix side of things, I will be returning to the Dirac Operator like I promised and what it has to do with the gravitational charge.

(Of course, reading this back it's like I am implying my matrices are traceless, which they are not... I was just throwing some idea's about).
Last edited by Simon Belmont on Thu Oct 11, 2012 1:53 pm, edited 1 time in total.

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### Re: Planck Particles

Motz' Work and my Interpretation of the Dirac Operator in terms of the Mass Charge

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.

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.

......

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

$\hbar = \frac{GM^2}{c}$
Last edited by Simon Belmont on Thu Oct 11, 2012 6:11 pm, edited 1 time in total.

Simon Belmont
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Joined: Fri Sep 14, 2012 10:58 am

### Re: Planck Particles

Simon Belmont wrote:Motz' Work and my Interpretation of the Dirac Operator in terms of the Mass Charge

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.

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.

......

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 can 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.

Simon Belmont
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### Re: Planck Particles

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 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

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

...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.

Simon Belmont
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### Re: Planck Particles

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.

Simon Belmont
Posts: 38
Joined: Fri Sep 14, 2012 10:58 am

### Re: Planck Particles

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, Compton.

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

$\hbar c = GM^2$