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Coupling Using Theta as a Boundary

  1. Jun 23, 2011 #1
    Suppose I have a mass [tex]M_0[/tex] (here denoted with lowercase zero because of previous discussions on relativistic mass), and I have a gravitational field [tex]\phi[/tex] which can under make a shift of [tex]180^o[/tex] 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:

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

    The question is the coupling. Since the boundary of the sum is the shift of [tex]-sin \theta[/tex] and [tex]-cos \theta[/tex] then [tex]\phi[/tex] is related to the mass by the probability coupling field [tex]\Psi[/tex]. Have I made my coupling correctly?
  2. jcsd
  3. Jun 24, 2011 #2
    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 [tex]M^2\phi^2[/tex]-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 [tex]\pi \in (\mathcal{R},\mathcal{C})[/tex]. 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'' [tex]\Delta \phi(x)[/tex], you find

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

    the interaction term [tex]M_{i}\psi_{i} \psi^{\dagger}_{i}[/tex] on [tex]\phi(\psi_i)[/tex] insures a self-interactive Hamiltonian [tex]\mathcal{H}[/tex]. 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 [tex]\frac{eh}{2Mc}[/tex].

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

    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 [tex]\phi=0[/tex] 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 [tex]\phi[/tex], that the electromagnetic interaction [tex]D_{\mu}D^{\mu} \phi[/tex] which has the value of [tex]M^2 \phi[/tex] -
    this is famously known as the mass-squared term [tex]M^2[/tex].
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