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

[Aethelwulf]

I really should simplify the equation which describes the Planck Particle in terms of the second order Dirac Operator

 

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

 

It's looks like a complicated mess... I have also decided to change the operator coefficient on the Planck time, usually the large Delta operator measures large changes, the simplified version of the entire equation should take the form

 

[math]D \psi(x) = \Delta E^{2}_{p} \delta t_{p}^{2} \frac{1}{24r_s^2}\psi(x)[/math]

Edited October 14, 2012 by Aethelwulf

[Aethelwulf]

I really should simplify the equation which describes the Planck Particle in terms of the second order Dirac Operator

 

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

 

It's looks like a complicated mess... I have also decided to change the operator coefficient on the Planck time, usually the large Delta operator measures large changes, the simplified version of the entire equation should take the form

 

[math]D \psi(x) = \Delta E^{2}_{p} \delta t_{p}^{2} \frac{1}{24r_s^2}\psi(x)[/math]

 

 

Well... that goes for it a clumsy mistake I made in the math, been so focused on certain parts of the equation, I've realized that it is not time independant, which means my wave function must also depend on time... whoops!

 

[math]D \psi(x,t) = \Delta E^{2}_{p} \delta t_{p}^{2} \frac{1}{24r_s^2}\psi(x,t)[/math]

Edited October 14, 2012 by Aethelwulf

[Aethelwulf]

Going back to the matrices, you can reajust some well-known equations due to the equivalent matrices I have presented.

 

We have already established that we can write the two dimensional wave function for the Dirac Operator in terms of Spinors, we obtain

 

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

 

For a three dimensional case, involves

 

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

 

The Spin of a Dirac Operator is therefore provided as

 

[math]D = -i \sigma_x \partial_x - i \sigma_y \partial_y[/math]

 

For Chirality in the modified formula, we can describe it as a matrix

 

[math]\Psi = \displaystyle \binom{\psi_R}{\psi_L}[/math]

 

where this is the same as saying

 

[math]\Psi = \displaystyle \binom{\frac{1 - \gamma^{5^{*}}}{2}}{\frac{1 + \gamma^{\hat{5}}}{2}}[/math]

 

This form describing Chirality is actually well-known, only that we have described [math]\gamma^5[/math] to suit my own equivalent matrices

 

[math]\gamma^{5^{*}} = - i\bar{\gamma}^{1_{2}} \bar{\gamma^{1_{2}}} \gamma^2 \gamma^3[/math]

 

[math]\gamma^{\hat{5}} = + i\bar{\gamma}^{1_{2}} \bar{\gamma^{1_{2}}} \gamma^2 \gamma^3[/math]

 

You could even go as far and calculate the angle between two spin vectors, I might do this at some point.

Edited October 14, 2012 by Aethelwulf

[Aethelwulf]

Just some more remarks on the equation...

 

[math]D \psi(x,t) = \Delta E^{2}_{p} \delta t_{p}^{2} \frac{1}{24r_s^2}\psi(x,t)[/math]

 

I need to be extra careful with my notation. You see, when I did my work originally extending Motz' idea to Planck Particles, he wrote the second order Dirac Operator as simply [math]D\psi(x)[/math]. Notation then seems to have varied since, because usually [math]D[/math] is the first order operator whilst [math]D^2[/math] would of course be it's second order.

 

Another... important point to make is this... time dependency found in the modified version of the Operator I give above. Normally [math]D^2 = \Delta[/math] where [math]\Delta[/math] is the Laplacian. The Laplacian is however, three dimensional case on the vector bundle [math]V[/math].

 

However, the Planck Particle is strictly speaking, a high-energy physics system. This means that we are allowed to relax the notation and let the second order Dirac Operator equal the Laplacian, but as I said, the Laplacian is in fact a three dimensional case, and the Operator proposed which can help describe these subatomic particles is in fact a four-dimensional case, it simply cannot be ignored that the right hand side has a time-dependency, so it seems only logical to assume the second order Dirac Operator in this case can be relaxed to the D'Alembertian, the four-dimensional but still equally dimensional consistent case of the three dimensional Laplacian.

Edited October 15, 2012 by Aethelwulf

[Aethelwulf]

Later on tomorrow, I will take us through why Dirac proposed the idea why a square root of the Laplacian came into his equations. Then we will see how this all comes back to the idea of how geometry is involved with Planck Particles... They are after all, the fundamental objects of geometry itself.

[Aethelwulf]

One begins with the Cauchy-Riemann equation

 

[math]\frac{\partial f}{\partial \bar{z}} = 0[/math]

 

with

 

[math]\frac{\partial}{\partial \bar{z}} = \frac{1}{2}(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y})[/math]

 

The reason for this has to do with set theory, but that's really irrelevant and I just want to get to the point.

 

The equations are considered on an [math]R^2[/math] flat Euclidean space, which has the possibility of being orientated. To do that you need to replace the operator

 

[math]\frac{\partial}{\partial \bar{z}}[/math]

 

with a differential operator

 

[math]\frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y})[/math]

 

Taking both the operators together we can obtain

 

[math]P \displaystyle \binom{f}{g} = 2i \begin{pmatrix} \frac{\partial g}{\partial z} \\ \frac{\partial f}{\partial \bar{z}} \end{pmatrix}[/math]

 

A calculation can lead to an alternative formula for [math]P[/math]

 

[math]P = \gamma_x \frac{\partial}{\partial x} + \gamma_y \frac{\partial}{\partial y}[/math]

 

where we are introducing the gamma matrices

 

[math]\gamma_x = \begin{pmatrix} 0 & i \\i & 0 \end{pmatrix}[/math]

 

and

 

[math]\gamma_y = \begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix}[/math]

 

and that

 

[math]\gamma^{2}_{x} = -E = \gamma^{2}_{y}[/math]

 

and also satisfying the algebra

 

[math]\gamma_x \gamma_y + \gamma_y \gamma_x = 0[/math]

 

The square of the Operator [math]P[/math] is the Laplacian on the two dimensional Riemannian surface

 

[math]P^2 = -\frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2}[/math]

 

which means simply that [math]\sqrt{P} = \Delta[/math]

 

Consider now the relativistic energy momentum relationship

 

[math]E = \sqrt{c^2p^2 + M^2c^4}[/math]

 

Canonically quantizing this equation by replacing the energy and momentum with their respective quantum operators, we can get

 

[math] i\hbar \partial_t \psi = \sqrt{c^2\hbar^2 \Delta + M^2 c^4}\psi [/math]

 

This is actually now a three dimensional case of a Laplacian, becayse the state function [math]\psi(t,x)[/math] can be defined as [math]R^1 \times R^3[/math]

 

In my own equation attempting to describe Planck Particles through such an operator, we are working explicitely in a four-dimensional case [math]R^4[/math], because of my speculations that we can relax the operator even further for the d'Alermbertian [math]\Box[/math] if the equation depends on time, which it clearly does since the temperature of these particles (and so the system itself) depends on the time allowed for these particles to exist.

 

ref.

 

http://www.maths.ed.ac.uk/~aar/papers/friedrich.pdf

Edited October 20, 2012 by Aethelwulf

[Aethelwulf]

To go from a three dimensional case to a four dimensional case however, one needs to think about the Minkowski metric - this is not impossible, in fact, one could argue the work had done itself since making the metric four dimensional simply means you are rotating via a Lorentz Transformation in a four dimensional Euclidean space.

[Aethelwulf]

Bridging the Gap between the Source of the Gravitational Field and the CODATA Charge Definition

 

Just to quickly summarize the previous work I said

 

The charge due to fundamental relationships is given as

 

[math]e = \sqrt{\frac{2\alpha \pi \hbar}{\mu_0 c}}[/math]

 

Therefore, because of the relationship we have already covered:

 

[math]e^2 = 4\pi \epsilon_0 GM^2 = 4 \pi \epsilon_0 \hbar c[/math]

 

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

 

[math]\sqrt{G}M = \sqrt{\frac{\pm \alpha \hbar n}{2 \epsilon_0 \mu_0 c}}[/math]

 

From here I asked ''is it possible the constants on the right somehow determine the source of the gravitational field?''

 

We established that the angular momentum was somehow conserved through the gravitational charge.

 

But how could it explain the source of the gravitational field? I came to some (possible) conclusions how it could be interpreted.

 

Permeability is the measure of the ability of a material system to support the formation of a magnetic field within itself and the permittivity is a measure of how an electric field affects and is also affected by a dielectric medium.

 

A particle like an electron, induces a magnetic field when it moves and magnetic fields contain energy, so a common question is where does the energy come from to create such a field?

 

It's come from the force which accelerated the electron from a (near rest state). This means that an electrons inertia comes from the supplying energy to the induced magnetic field (1)

 

Inertia is however, strictly a property of mass itself according to the modern theory of Einstein's relativity, which would also mean it is a property of the gravitational charge.

 

This is the first realization of it's kind concerning the innert gravitational charge of a system and the inertia contained therein of the system in question when regarding an induced magnetic field.

 

This may even bring about questions whether the inertia of particles is due to their energy content as once speculated by Einstein and was raised very early on in my work in the beginning and may also bring us about to the idea that there is actually an electromagnetic mass for particles.

 

 

(1) http://www.mariner.connectfree.co.uk/html/e_m_inertia.html

[Aethelwulf]

However, if we are talking about the energy of inertia, for an electron induced by a magnetic field, how does the energy come into the equation

 

[math]\sqrt{G}M = \sqrt{\frac{\pm \alpha \hbar n}{2 \epsilon_0 \mu_0 c}}[/math]

 

Well that's actually simple. Going back to my derivations, I found the relationship

 

[math]\sqrt{Er_s} = \sqrt{G}M[/math]

 

This means energy appears in the equation as

 

[math]Er_s = \frac{\pm \alpha \hbar n}{2 \epsilon_0 \mu_0 c}[/math]

 

Where [math]r_s[/math] is the Schwarzschild radius again.

[Aethelwulf]

So... what is the next step? If we are talking about electromagnetic mass, the question should be, ''how much of the self-energy of the particle is contributing to the gravitational charge (mass)?''

 

The solution may come interestingly from the gravitational analogue of the elementary charge - Remember, [math]e^2[/math] is the same thing as saying [math]4 \pi \epsilon GM^2[/math] where [math]GM^2[/math] is the squared gravitational charge. One may have a gravtostatic equation of contribution of energy to mass by

 

[math]E_g = \frac{1}{2} \frac{4 \pi \epsilon GM^2}{r_s}[/math]

 

Where [math]E_g[/math] denotes our energy contributed to the intrinsic gravitational charge. One may also notice this is the gravitational analogue of

 

[math]E_{EM} = \frac{1}{2} \frac{e^2}{R_{classical}}[/math]

 

The contribution of mass in my equation is found then as

 

[math]M = \frac{1}{2} \frac{4 \pi \epsilon GM^2}{r_s c^2}[/math]

Edited November 5, 2012 by Aethelwulf

[Aethelwulf]

Using a similar process to Wein (1900), then the attraction of the gravitational field can be understood as the squared gravitational charge divided by the distance

 

[math]G \frac{\frac{1}{2} \frac{4 \pi \epsilon GM^2}{r_s c^2} M}{R}[/math]

 

Where [math]r_s < R[/math] as [math]R[/math] spans the distance between the charges [math]M_1[/math] and [math]M_2[/math]

Edited November 5, 2012 by Aethelwulf

[Aethelwulf]

And to finish that off, the gravitoelectromagnetic equation (that is) both the electromagnetic contribution and that which is intrinsically contributed by the gravitational aspect appears as:

 

[math]E_{GEM} = \frac{1}{2}(\frac{4 \pi \epsilon GM^2}{r_s} + \frac{e^2}{r_s})[/math]

 

Where [math]r_s[/math] is assumed smaller than the classical radius [math]\frac{e^2}{Mc^2}[/math] (but still a sphere, and [math]e^2[/math] plays the role of the electric charge.

Edited November 5, 2012 by Aethelwulf

[Aethelwulf]

Sorry, I meant ''gravitoelectromagnetic.''

Edited November 5, 2012 by Aethelwulf

[Chewbalka]

This all appears to be crazy talk to me! I never understand these formulas because I never know the values of the numbers as well as I have never been any good at math. I tried figuring out why I consistantly got the wrong answer on a previous post on thermal expansion but eventually just gave up... I was only off by a small amount but that little bit I am sure adds up when it needs to be accurate... I love physics but I need a giant cheat sheet just to get by lol. Well I am not sure what it is your doing on this topic but I am sure its smart guy stuff and it looks like ya got a good grip on it lol so ill leave it be!

[Aethelwulf]

And to finish that off, the gravitoelectromagnetic equation (that is) both the electromagnetic contribution and that which is intrinsically contributed by an electric charge appears as:

 

[math]E_{GEM} = \frac{1}{2}(\frac{4 \pi \epsilon GM^2}{r_s} + \frac{e^2}{r_s})[/math]

 

Where [math]r_s[/math] is assumed smaller than the classical radius [math]\frac{e^2}{Mc^2}[/math] (but still a sphere, and [math]e^2[/math] plays the role of the electric charge.

 

This solves the [math]\frac{4}{3}[/math] mass problem - If and only if the contribution [math]\frac{4 \pi \epsilon GM^2}{r_s}[/math] plays the role of the Poincare Stresses - In essence, we must assume that [math]G[/math] takes on extremely large values inside of the sphere which help cancel the electrostatic forces believed to rip a particle apart. We can assume [math]G[/math] takes on an extremely large value by going back to Motz paper where he once admitted it having a large value through

 

[math]\hbar c = GM^2[/math]

 

via

 

[math]G = \frac{\hbar c}{M^2}[/math]

 

In which he states that the gravitational constant is taken to be very large inside the sphere of a particle. Adopting this equation then allows us to ignore Poincare Stresses. but does raise an important question of fine tuning.

Edited November 5, 2012 by Aethelwulf

[Aethelwulf]

It can play the role of the Poincare Stresses because it plays a non-electromagnetic energy, which he defined as [math]E_p[/math].

[Aethelwulf]

Just a small reference on what Poincare Stresses are. They are not often spoke about in physics any more...

 

http://www.answers.com/topic/poincare-stresses

 

One reason they are not spoke about is because the scientific community leans towards particles are truly point like. This is heavily argued in my work.