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 $$\mathbf{\Phi}$$

$$ (1) \mathbf{\Phi}=(\Phi_0,\vec{\Phi})=\left(\Phi_0, \Phi_x, \Phi_y, \Phi_z \right)=\left(-E, k_x, k_y, k_z\right)\exp\left(i(Et-k_x x-k_yy-k_zz)\right) $$

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.