Massless particles and the Neutrino

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 $\mathrm{\Phi }$

$(1)\mathrm{\Phi }=({\mathrm{\Phi }}_0,\overrightarrow{\mathrm{\Phi }})=({\mathrm{\Phi }}_0,{\mathrm{\Phi }}_x,{\mathrm{\Phi }}_y,{\mathrm{\Phi }}_z)=(-E,k_x,k_y,k_z)\exp (i(Et-k_{x}x-k_{y}y-k_{z}z))$

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

With (1) we obtain $(2){\partial }_t{\mathrm{\Phi }}_0=\overrightarrow{\mathrm{\nabla }}\overrightarrow{\mathrm{\Phi }}$

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 $\varphi$ can be defined by

$(3)\varphi =(1,s_x,s_y,s_z)\exp (i(-Et+k_{x}x+k_{y}y+k_{z}z))$

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 $(4){\partial }_{t}1\exp (i(-Et+k_{x}x+k_{y}y+k_{z}z))=-(s_x,s_y,s_z)\overrightarrow{\mathrm{\nabla }}\exp (i(-Et+k_{x}x+k_{y}y+k_{z}z))$

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

$(5)i\hslash \frac{\partial \psi }{\partial t}=[c\overrightarrow{\alpha }(\overrightarrow{p}-m_0\overrightarrow{\mathrm{\Phi }})+m_0{c}^2{\mathrm{\Phi }}^0+c^2\beta m_0]\psi$

to the propagation of the photon

$(6)\frac{\partial \exp (i(-E_{photon}t+k_{x}x+k_{y}y+k_{z}z))}{\partial t}{s}_0=[(\overrightarrow{p}-\frac{E_{photon}}{c^2}\overrightarrow{\mathrm{\Phi }})(s_x,s_y,s_z)+E_{photon}{\mathrm{\Phi }}^0]\exp (i(-E_{photon}t+k_{x}x+k_{y}y+k_{z}z))$

When we suppose that $\overrightarrow{\mathrm{\Phi }}$ 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

$(7)E_{photon}=k_z(1+{\mathrm{\Phi }}^0)$

and thus for the respective energies of the photon at positions $z_1$ and $z_2$ an energy difference $\mathrm{\Delta }E$

$(8)\mathrm{\Delta }E={\mathrm{\Phi }}^0(z_1)-{\mathrm{\Phi }}^0(z_2)$

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