Massless particles and the Neutrino

Resource type: this resource may consist of Fringe science.

This page is under construction by Hartwig Poth, hupoth@gmail.com

For the four potential theory of gravitation, cf. Four potential theory of gravitation, a quantum particle, the 'gravon', can be found. It is massless and is defined by the following four vector wave function $\mathbf {\Phi }$

$(1)~~~\mathbf {\Phi } =(\Phi _{0},{\vec {\Phi }})=\left(\Phi _{0},\Phi _{x},\Phi _{y},\Phi _{z}\right)=\left(-E,k_{x},k_{y},k_{z}\right)\exp \left(i(Et-k_{x}x-k_{y}y-k_{z}z)\right)$

wherein $\mathbf {\Phi }$ is the gravitational four potential . The gravon has obviously the spin 0.

With (1) we obtain $(2)~~~\partial _{t}\Phi _{0}={\vec {\nabla }}{\vec {\Phi }}$

This is a linear wave equation for the gravon and also a continuity equation; it is furthermore the third Maxwell type field equation for the four potential gravitation.

In analogy to (1) a photon $\phi$ can be defined by

$(3)~~~\phi =(1,s_{x},s_{y},s_{z})\exp \left(i(-Et+k_{x}x+k_{y}y+k_{z}z)\right)$

wherein $(s_{x},s_{y},s_{z})$ is the spin vector for the spin 1 of the photon; the spin vector behaves like an ordinary spatial vector and is directed along the momentum $(k_{x},k_{y},k_{z})$ of the photon. The time like component $s_{0}$ of the complete spinor $(1,s_{x},s_{y},s_{z})$ in (3) is $1$. The linear wave equation for this photon is $(4)~~~\partial _{t}1\exp \left(i(-Et+k_{x}x+k_{y}y+k_{z}z)\right)=-(s_{x},s_{y},s_{z}){\vec {\nabla }}\exp \left(i(-Et+k_{x}x+k_{y}y+k_{z}z)\right)$

That equation (4) allows for example to apply the four potential of gravitation $(\Phi ^{0},{\vec {\Phi }})$ in analogy to the influence of the common electromagnetic four potential onto the Dirac electron

$(5)~~~i\,\hbar {\frac {\partial \psi }{\partial t}}=\left[c\,{\vec {\alpha }}\,\left({\vec {p}}-m_{0}\,{\vec {\Phi }}\,\right)+m_{0}\,c^{2}\,\Phi ^{0}+\,c^{2}\beta \,m_{0}\right]\psi$

to the propagation of the photon

$(6)~~~{\frac {\partial \exp \left(i(-E_{photon}t+k_{x}x+k_{y}y+k_{z}z)\right)}{\partial t}}s_{0}=\left[\left({\vec {p}}-{\frac {E_{photon}}{c^{2}}}\,{\vec {\Phi }}\,\right)(s_{x},s_{y},s_{z})+E_{photon}\,\Phi ^{0}\right]\exp \left(i(-E_{photon}t+k_{x}x+k_{y}y+k_{z}z)\right)$

When we suppose that ${\vec {\Phi }}$ vanishes, if the source of gravitation is virtually at rest and if we consider for simplicity a photon which propagates along the $z$-axis with the momentum $k_{z}$ we obtain eventually

$(7)~~~E_{photon}=k_{z}(1+\Phi ^{0})$

and thus for the respective energies of the photon at positions $z_{1}$ and $z_{2}$ an energy difference $\Delta E$

$(8)~~~\Delta E=\Phi ^{0}(z_{1})-\Phi ^{0}(z_{2})$

This result is already known from the general theory of relativity .