Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.

Astro · 2010-07-28 00:13:47

This is the formulation of the gravitational charge unification in a model to satisfy inverse relationships with electric charge. It seems such a unification can explain how energy can flux into matter, taking in several more accounts to explain why.

$\vec{F}= q \phi g= (-\nabla \phi \frac{E\psi}{c^2})$

The orthogonal state we began with, is analougous to perturbation of its equality $|\phi_n (t)>$ where the subscript $n$ dictates the energy level. Applying this to our constant force in the equation

$\vec{F}= (-\nabla \phi \frac{i\hbar \partial_t}{c^2})$

then shouldn't the n'th permutations in the system follow such an equation|?

$\vec{F}_{n}|\phi(t)> = (-\nabla \phi \frac{i\hbar \partial_t}{c^2})(e\frac{-i \omega_nt}{|\phi>})$

This approach led me to: This part is set on mostly postulations, however, whilst extremal objects can be used in order to help understand perturbation properly, it seems that a well-known component can become as a restraining system.

Near-extremal black holes are usually performed on by using an energy-perturbation theory around the extremal black hole/singularity/particle, which will be the mathematical foundation of the following theories.; psuedoscientific theories, unfortunately… but still possible.

Some theories involving additional space dimensions predict that micro black holes could be formed at an energy as low as the TeV range. No doomsday scenario can be found though, because the extremity of this singularity cannot gobble matter, no more than it acts like a black hole which cannot emit radiation.

$g_{\mu \nu}=g^{0}_{\mu \nu} + \hbar_{\mu \nu}$

For a spherical body which is not point like on a symmetric metric tensor, and so describes the energy associated to that perturbed singular particle with a structure. The small perturbations reside in the form of $\hbar_\nu \mu$ which then must be analyzed carefully.

And so, perturbations, if i did the math properly would lead to ~

$g_\mu \nu=g^{0}_{\mu \nu} + \hbar_{\nu \mu}$

If we make a bold assertion that the metric itself $g_\mu \nu$ is equal to that of the Hamiltonian $\hat{H}|\psi>$ , then we can by theory perturbate $\hbar_\nu \mu$ in the orthogonality of state around the singular structure, by the identity of relations $|\phi_n>$ . The reason why, and how, will come soon.

From similar gravitational relativistic equations, we have for a Langrangian: $S=\int L \sqrt{-g} d^4 x$ which satisfies a structural system for a particle because it has a density which can only be attributed to a volume $\rho= \nabla^2 \phi$ , thus the curvature is the mass and the structure is the proposition that can be obtained, even in a Least Action of Density.

For a spherical body which is not point like on a symmetric metric tensor, and so describes the energy associated to that perturbed singular particle with a structure. The small perturbations reside in the form of $\hbar_\nu \mu$ which then must be analyzed carefully.

These following derivations are the only way I can describe such a theory in terms of geometrical relativity:

$i\hbar \partial_t |\psi >=\hat{H}|\psi>$

And knowing $g_{\mu \nu}=g^{0}_{\mu \nu} + \hbar_{\nu \mu}$

Then my relation which is quite an assumption is $g_{\mu \nu}|\psi> = \hat{H}|\psi>$ , just think of it as the energy of our syatem which in this case, will follow the principle of least action.
If this is true, then it follows the perturbations of Orthogonality States $\phi>$ , so calculating it in terms of a minimalizing integral:

$\delta \int \hat{H}|g_{\mu \nu} |\psi> = g^{0}_{\mu \nu} |\psi> + \hbar_{\mu \nu}_{0}|\psi>$

There are many meanings here. It means that the perturbed state of $g_{\mu \nu}$ on a singularity/particle in spacetime is expressed as the ground state of $\hbar_{\mu \nu}_{0}|\psi>$ , which just states that the perturbation are as small as they can get, and $g^{0}_{\mu \nu}$ escalate to produce the lowest energy, but is the reason why $\hat{H}|\psi>$ for $g_{\mu \nu}$ is small, is because of a limiting condition found in Fermi-Energy Levels.

The final equation if I have done this right, should state:

$\delta \int <\psi|E_n+g_{\mu \nu}|\psi_n> = <\phi_n|g^{0}_{\mu \nu} + \hbar^{\mu \nu}_{0}|\phi_n>$

The equations of perturbation will require a final description of an inertial charge-mass relationship, and I derived one i think would suit, as simple as it is:

$I_M=E \frac{G}{c^2}$ where $I_M= \mu$

If left is defined as an inertial matter, then there must be an inertial energy (WHICH will be shown soon as being equivalent to the energy sqaured boundary.

The gravitational charge corresponds to $\mu_g$ . From here, we find solutions by algebraic manipulations:
$I_M=GM=E(G/c^2)=\mu$ thus multiply the distance out and then simplify for:

$I_M d = E \frac{G}{c^2}_0 \frac{1}{2} a t^2$

Taking the potential we have

$I_{M} dg =e^{i}\partial_{i} (E {G}{c^2}_0 ) \frac{1}{2} a t^2$

Using an Einstein summation for $-\nabla \phi$

Simplifying $I_{M} dg= e^{i}\partial_{i} {E \frac{G}{c^2}_0 \frac{1}{2} a t^2$

Leads to $I_{M} dg= (-\nabla \phi) \mu_g \frac{1}{2} at^2$

The charge located on the left can be identified specically with an inertial mass.

Where here the final equation is only important.

I did however come to note, that if the Luxon Hypothesis was to be satisfied, then the following must be held true:

In general relativity, the gravitational potential is provided by a metric tensor $g_{\mu \nu}$ - it is itself a 4x4 matrix. It's only when the gravitational potential reaches the components of the metric with $00$ as the index of $g$ does it couple to some external energy density.

Perturbing the state of the gravitational field aka. the Newtonian Potential $\phi$ , we would normally have:

$g_{00} = (1+2 \phi)$

This equation hopefully should have some major importances with my equation of the Gravi-gyro charge parameter, through the recognition of the Newtonian Gravitational Parameter.

But the final aim became clearer as i analyzed this approach. If one takes the notion that $M=q \phi$ then it seems that the electric charge can contribute to the gravitational charge, and since $M$ alone can be seen as not only inversely-related to the gravitational gradient of the system $\phi$ , then it can alone itself be considered the gravitational charge, even though i have come to promounce it in a series of equations as $\mu$ .

Thus, using a seperate equation that helps describe this better, i came to realize that:

$\mu^2_g= (-\nabla \phi \frac{-\hbar \partial_t}{c^2}q)\frac{G}{c^2}$

,then we have an electric charge relation to the gravitational charge located on the left of the equation.

This equation is analogous in the sense of $M=q \phi$ . So perhaps the inverse relationships are much more important than i first gave credit, thinking they were wrong.

The reason why?

The $g_{00}$ component of gravitational potential energy in the metric, will also depend on the restraint of the equation

$\delta \int <\psi_n|E_n+g_{\mu \nu}|\psi_n> = <\phi_n|g^{0}_{\mu \nu} + \hbar^{\mu \nu}_{0}|\phi_n>$

given earlier, following a principle of least action.

If the energy of the system follows such rules, then its dimensionally consistent for the Hamiltonian to be expressed as:

$\delta \int eE \mathbf{L}|g_{\mu \nu} |\psi> = g^{0}_{\mu \nu} |\psi> + \hbar_{\mu \nu}_{0}|\psi>$

Now, this final equation helps to retain very small perturbations $\epsilon$ which is the ''small'' measurements made on the restraint of $\hbar_{\mu \nu}_{0}|\psi>$ which has been summed over the metric component. This is the Generalistic Approach to treating energies of a system in the most efficient way possible, given the eignestate conditions the system is under, which may involve such observables like spin, velocity and even their internal counterparts.

The notation $\mathbf{L}$ describes the energy of the system (which in the presence of the electric charge $e$ or given as $q$ before) $eE$ means this is a system of energy which is an ionized particle. The ionization must play a fundamental role in the presence of matter, if tangibility of photon energy in these perturbative states are to be taken seriously to satisfy a unification of the gravitational charge as being a same manifestation of the presence of an electric charge in the equation:

$\mu^2_g= (-\nabla \phi \frac{-\hbar \partial_t}{c^2}q)\frac{G}{c^2}$

This means that a presence of a gravitational charge must be satisfied when the energy of the system $-\hbar \partial_t$ is multiplied by the presenc of the electric charge $q$ and the gravitational potential $(-\nabla \phi)$ which will satisfy the gradient of how much mass is obtained from the perturbation. This ultimately means that the Hierarchy model can be answered by initializing certain energy wavelengths of photons under close examination of what particles the produce. The final gravitational charge will depend on those energies which must be able to be calculated under the scrutiny of experimentation.

Last edited by Astro (2010-07-28 12:13:05)

Astro · 2010-07-28 02:34:27

Moving onto something else, i wanted to take into account a new identity given as:

$e^2 \phi (-\hbar \partial_t \mathbf{L})= \frac{1}{2}(\frac{\hbar \partial_t}{c^2}v)(\frac{-\hbar}{i} \nabla)$

Where we have used the energy and momentum operator to quantize the equation. This was obtained by manipulating the equations $M=q \phi$ and $eE \mathbf{L}= \frac{1}{2}Mv^2$ .

Messing around with one of the sides in the ideal nature of treated the gravitational field as an $n^{th}$ permutation, i came to the expression of

$\sum_n e^2 \phi_n(i \hbar \partial_t \mathbf{L})$

Since in this expression the gravitational field is the only permutated variable, then the energy can be considered constant in this case, and as we will find, will not depend at all on the permutations of the field.

The expression can be reconfigured to give:

$e^2 \phi (i \hbar \partial_t \mathbf{L}) \sum_n \nabla \phi_n$

Notice as well we have taken the correct notation for the gradient of the gravitational field potential. It is very important now to note that there is a restraint on the equation if we take the curl of rotational infinitessimal density of the torsional gravitational field.

$e^2 \phi (i \hbar \partial_t \mathbf{L}) \sum_n \nabla X (\nabla \phi_n) = 0$

This restraint is important because if the systems energy is gained when moving a distance $\mathbf{L}$ , then does not necessrily gain this energy from the gravitational component of the field:

$\frac{1}{2}(\frac{i \hbar \partial_t}{c^2}v)(\frac{-\hbar}{i} \nabla) + \sum_n \nabla X (\nabla \phi_n) = e^2 \phi (-\hbar \frac{\partial}{\partial t} \mathbf{L)}$

Which means quite clearly that the addition of the permutations of the curl of the torsional field in terms of graviation does not add anything to the overall energy gained from the system when moving some distance.

Last edited by Astro (2010-08-01 00:05:30)

Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.

#1 2010-07-28 00:13:47

Luxon Hypothesis

#2 2010-07-28 02:34:27

Re: Luxon Hypothesis

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