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

Goldstone · 2011-06-23 18:53:23

I would have wrote the whole paper I made, but I thought it would be best to just keep to mathematical side: what needs to be known is that mass is a charge, usually seen as the Charge of a Higgs Field. Mass is also seen as a charge in the equation $F_g= -\nabla \phi Mg$ (see Sean Carrols notes on Relativity), mass can also be a charge of the Gravitational field, which can be seen in the simple energy system $M\phi$ . The square of this quantity leads to an oscillation of energy in QFT. Using similar idea's on the energy-mass equation based on Einsteins relativity, a similar equation is formed and is seen in light of a geometry of the field and the energy density which makes it.

$\phi = \phi_R + \phi_C$

has a real and complex part, so our rotations will be made through a complex and a real range $\pi \in (\mathcal{R},\mathcal{C})$ . We can see this best as:

$\phi = e^{isin(\theta + \pi)}\phi$

$\phi' = e^{-icos(\theta +\pi)}\phi'$

So we have two solutions to the rest energy of relativity:

$E=M(e^{isin(\theta + \pi)})\phi$

$E=M(e^{-icos(\theta + \pi)})\phi'$

If the field was a gradient energy, then it is contributing to the Hamiltonian of this system. If this is the case, we can see that:

$E=M\Delta\phi(x)$

since

$\Delta \phi(x) =\phi(x) \rightarrow \phi*(x)$

which through substitution, gives a more simplistic form

$\Delta E=M\phi(\Lambda^{-1} x)$

There is also now the question of what energy has been taken by the shift. If the change in the Hamiltonian of the system was purely gravitational and no other added energies, then the shift mathematically can be given as:

$\phi(x) \rightarrow \phi*(x)=\phi(\Lambda^{-1} x)$

where the inverse $\Lambda^{-1}$ states that the field has definately been shifted. If $\phi$ is no longer arbitrary, we can state it is actually the gravitational field, then:

$\Delta E=M\phi(\Lambda^{-1} x)$

So the field shift in our energy $M\Delta \phi(x)$ I argue is a contribution of the energy contained within the local field of the gravitational field on something like a photon. Instead of wanting to calculate broken symmetries to explain mass (and then of course a Higgs Boson and a superfluous field) you end up with mass being described under two phases experienced in the field. One field predicts quantum energy, and the other mass, a more concentrated form of the same thing.

Mass is then determined by some negative and positive solution due to energy shifts in the field $\phi$ . But the coupling has not been expressed properly; it really requires the use of a probability field $\Psi$ which couples the mass to the dynamics of the $\phi$ field:

$\Delta E \Psi= \sum_{i}^{\theta} M_{0i} \phi (\Lambda^{-1} x) \psi_i$

The Hilbert space is also within the complex and real ranges, so it does not invoke unitary solutions in every case:

$\phi \in (\mathcal{R},\mathcal{C}) \to \mathcal{H} \in (\mathcal{R},\mathcal{C}) \forall \psi(\theta)$

The $\phi$ field is in the complex and real range which directly means that the Hilbert space is also within the same ranges, for all changes in the energy of the fields shift which we can dictate as part of a wave function as we have seen.

Last edited by Goldstone (2011-06-29 01:52:56)

Goldstone · 2011-06-23 21:14:28

I want to expand on the usefulness of the equation:

$\Delta E \Psi= \sum_{i}^{\theta} M_{0i} \phi (\Lambda^{-1} x) \psi_i$

It predicts charge changes in particles, it could be used for the Dirac Equation as an explanation for right moving waves and left moving waves. It can predict the energy perturbations in local mass, meaning that when $\gamma \gamma \rightarrow M^{-}M^{+}$ so that the energy implies the creation of mass, the equation gives it meaning that the mass created was transferred from the gravitational field. The equation is source field equation also, just as a mathematical side track. The equation is unique in that it is a special case of the $U(1)$ symmetry group. The symmetry exists in the phases, but the phases have negative solutions so it cannot be unitary all the time. It could then explain oscillations of the field ''zitter - motion'' then arises as being the mass of the particle, which is the charge of the field, self-interacting on the field. It is said that on global scales, the gravitational field is non-self-interacting. It may interact on local scales of particles however as an energy gradient in the gravitational field - which should hold potential energy for particles if we are to believe the postulates of General Relativity. All we are attempting to do is quantize the idea of the mass of the field, by saying the mass is the charge of the field, much in the same sense we assume easily the electron has a charge when moving in a field... So does a concentrated local energy have a probability field dictating when mass comes into play?

Last edited by Goldstone (2011-06-23 21:23:41)

Goldstone · 2011-06-23 21:19:03

extra copy, sorry. Deleted.

Last edited by Goldstone (2011-06-23 21:20:14)

Goldstone · 2011-06-23 21:42:35

In compact notation, our equations are in simplified form:

$\Delta E \Psi = \sum_{i}^{\theta} M_i (e^{isin(\theta + \pi)})\phi_i \psi_i$

$-\Delta E \Psi = -\sum_{i}^{\theta} M_i (e^{-icos(\theta + \pi)})\phi_{i}^{'} \psi_{i}^{\dagger}$

In a Dirac equation, the mass term couples left moving and right moving waves, so the mass term here has internal dynamics-relations with $(e^{isin(\theta + \pi)})$ because the value of M depends on the probability of the coupling field. So maybe in a sense, since the gravitational energy gradient holds two solutions which are positive and negative may be a reason in itself why psuedoscalar fields have not been identified. The mass in these two equations in this post play their part independant of the other equation, all the mass depends on is the scale factor $\phi$ . It then can be assumed or postulated that the two solutions should not meld to form zero-spin particles, so a mass term acting on the matrix of the Dirac Equation $\beta$ interchanges the sign of the particle can not be a real physical solution to reality. In the equations interpretation, they do not meld because the field shift in the negative plane holds a different range to the positive field shift, as would be expected.

Last edited by Goldstone (2011-07-06 08:15:15)

Goldstone · 2011-06-23 22:22:05

Interestingly, as I speak of the $M^2\phi^2$ -term as an oscillation of some field energy [1], I must be saying the square root of this energy is the range in which the field invariant takes on shift values of $\pi \in (\mathcal{R},\mathcal{C})$ . This is a gravitational charge energy - the mass of the quantum in question. Because one can invoke the inverse solution equation ''implying a change in the field'' $\Delta \phi(x)$ , you find

$\Delta E \Psi = \sum_{i}^{\theta} M_i \phi(\Lambda^{-1}x) \psi_i$

the interaction term $M_{i}\psi_{i} \psi^{\dagger}_{i}$ on $\phi(\psi_i)$ insures a self-interactive Hamiltonian $\mathcal{H}$ .

[1] - It plays a form, increased by one power of a matter term $M$ , that the electromagnetic interaction $D_{\mu}D^{\mu} \phi$ which has the value of $M^2 \phi$ - this is famously known as the mass-squared term $M^2$ .

Last edited by Goldstone (2011-06-28 23:56:04)

Goldstone · 2011-06-23 22:53:54

edited

Last edited by Goldstone (2011-06-28 23:57:06)

Goldstone · 2011-06-23 23:17:24

edited

Last edited by Goldstone (2011-06-28 23:56:37)

Goldstone · 2011-06-23 23:19:32

copied post again. Apologize.

Last edited by Goldstone (2011-06-23 23:21:32)

Goldstone · 2011-06-23 23:31:07

deleted

Last edited by Goldstone (2011-06-25 06:51:02)

Goldstone · 2011-06-29 00:14:44

In principle you can create particles with potential energy, the prototypical case is color confinement. If such cases exist, it is argued that potential energy can indeed contribute to the rest energy of a particle. The effects will remain local of course on the particle itself, and would imply that the potential field interacts with the structure of the particle. If indeed you can create particles from a potential field, then there is the suggestive thought that the potential has the dynamics in understanding negative and positive signs for mass charge.

Goldstone · 2011-06-29 00:33:31

So in order to explain the mass as a charge from the potential, you need a field density which puts the rest energy into a local confined space (for ever how short a time period that is due to Uncertain effects). If $g$ was a metric of relativistic freedom, then it's role in $\phi$ would be given as something like $\rho_{L} \approx g(M \phi_{L})$ which is the field density which is localized.

Goldstone · 2011-06-29 00:56:15

In the light of knowing this is an investigation into the confinement potential, knowing that mass is given in terms of a potential field, we can see in principle that mass is effectively part of the gravitational potential energy. Here in this link, it elaborates on the potential for quarks confinement potential gradients

http://143.50.77.55/itp/iutp/iutp_10/talks/popovici.pdf

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

#1 2011-06-23 18:53:23

Mass as as Charge from the Gravitational Field

#2 2011-06-23 21:14:28

Re: Mass as as Charge from the Gravitational Field

#3 2011-06-23 21:19:03

Re: Mass as as Charge from the Gravitational Field

#4 2011-06-23 21:42:35

Re: Mass as as Charge from the Gravitational Field

#5 2011-06-23 22:22:05

Re: Mass as as Charge from the Gravitational Field

#6 2011-06-23 22:53:54

Re: Mass as as Charge from the Gravitational Field

#7 2011-06-23 23:17:24

Re: Mass as as Charge from the Gravitational Field

#8 2011-06-23 23:19:32

Re: Mass as as Charge from the Gravitational Field

#9 2011-06-23 23:31:07

Re: Mass as as Charge from the Gravitational Field

#10 2011-06-29 00:14:44

Re: Mass as as Charge from the Gravitational Field

#11 2011-06-29 00:33:31

Re: Mass as as Charge from the Gravitational Field

#12 2011-06-29 00:56:15

Re: Mass as as Charge from the Gravitational Field

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