**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 $$e$$.

http://wdxy.hubu.edu.cn/ddlx/UpLoadF...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.

**The Radius**

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:

http://articles.adsabs.harvard.edu/full ... .113...34S

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)