# Coupling Using Theta as a Boundary

1. Jun 23, 2011

### Goldstone1

Suppose I have a mass $$M_0$$ (here denoted with lowercase zero because of previous discussions on relativistic mass), and I have a gravitational field $$\phi$$ which can under make a shift of $$180^o$$ between a negative plane and a positive plane. Assume also that the mass is considered as a charge, rather than something being seperate to it, and then:

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

The question is the coupling. Since the boundary of the sum is the shift of $$-sin \theta$$ and $$-cos \theta$$ then $$\phi$$ is related to the mass by the probability coupling field $$\Psi$$. Have I made my coupling correctly?

2. Jun 24, 2011

### Goldstone1

I haven't heard anything on my post. I wonder if that was because of a lack of information?

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 $$\mathcal{H}$$. What we have is a gravitational energy Hamiltonian that can be converted in the understanding of conventional mass weighing systems, There are many contributions to the system which does not involve the mass alone - for instance the energy of the electric and magnetic vacuum are not taken into consideration, but it will be a project of mine these next few weeks to understand that kind of system. For instance, as I have speculated, physics says that an electron absorption of a fluctuation of the zero-point energies has dimensions $$\frac{eh}{2Mc}$$.

$$\mathcal{H} \Psi = (\sum_{i}^{\theta} M_i \phi(\Lambda^{-1}x) + \frac{eh}{2Mc}) \psi_i$$

This equation is the total energy of the gravitational field with a contribution of zero-point energy, a photon in this case. So this is a quantum description of the energy of a field, the gravitational field - and the field occupies gradients where the photon in the zero-point field lie on the range $$\phi=0$$ as a ground state boundary system. This is just quantum mechanics right? I don't see how you can falsify the condition of the Hamiltonian...

[1] - It plays a form, increased by one scalar field $$\phi$$, 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 $$M^2$$.