Mystery111 Posted October 5, 2011 Posted October 5, 2011 (edited) I propose [latex]\bar{\psi}c(\gamma^i \cdot \hat{p})\chi + \bar{\psi}(\Box \phi \ell^3) c^2\psi = \bar{\psi} \gamma^0 (i\hbar \partial_t) \psi[/latex] where the rest part is derived from the mass dependance of the gravitational field [latex]\phi[/latex] in four dimensions. This an analogue of the dirac equation in graviattional interaction terms. The part [latex]\gamma^i[/latex] determines whether to express [latex]\hat{p}[/latex] as either positively or negatively and for rest energy particles travelling lower than the speed of light is given as [latex]\bar{\psi}c(\gamma^i \cdot \hat{p})\chi > c[/latex] (rest) This component [latex]\bar{\psi}(\Box \phi \ell^3) c^2\psi[/latex] is the intrinsic rest energy. It is non-relativistic on it's own. Of course, the full equation [latex]\bar{\psi}c(\gamma^i \cdot \hat{p})\chi + \bar{\psi}(\Box \phi \ell^3) c^2\psi = \bar{\psi} \gamma^0 (i\hbar \partial_t) \psi[/latex] Makes up the Hamiltonian for a free particle when relativistic momentum effects are reduced. If we take normal mathetical convention we have [latex]f=0[/latex] [latex]\bar{\psi}(\Box \phi \ell^3) c^2\psi = \bar{\psi} \gamma^0 (i\hbar \partial_t) \psi[/latex] If [latex]f[/latex] has dimensions of mass, then we can see that if [latex]\bar{\psi}c(\gamma^i \cdot \hat{p})\chi = fc^2\bar{\psi}\psi[/latex] You can insert some yukawa couplings in here, expand this equation out in consideration of a Higgs field with terms satisfying: [latex]g_{\gamma} \bar{\psi}\psi = g_{\gamma}f(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L) + g_{\gamma}H(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L)[/latex] the rest energy actually depends variationally on the state of the momentum component in the sense that the rest energy part requires the idea that mass requires the dependance on the gravitational field as Nordstrom classically proved. [latex]g_{\gamma}\bar{\psi}(\Box \phi \ell^3) c^2\psi =g_{\gamma}(\psi^{\dagger}_L \psi_R + f\psi^{\dagger}_R \psi_L) + H(\psi^{\dagger}_L \psi_R + \psi^{\dagger}_R \psi_L) )[/latex] This would mean that mass is a relativistic phenomenon; then that would mean that the momentum which directly influences variantional energy in the rest energy is provided to the system. It becomes covariant under the same translation you would give a fermion with a mass under the symmetry breaking of a [latex]U(1)[/latex] group. Edited October 6, 2011 by Mystery111
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