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3.5.4 Particle Energy and Associated Constants

Particle energy is related to the mass of the particle by the Einstein formula, \(E=Mc^2\), which can be verified by the following calculations.

It has been shown in the chapter on gravity that the displacement constant, \(A\), is proportional to the mass of the particle. Specifically,
\begin{align} A=\frac{G}{H\ c_0} Mass \nonumber \end{align}
where the Hubble constant, \(H=2.6436459\ 10^{-18}\),
the expansion velocity, \(c_0=2.99792\ 10^8\),
and the universal gravitational constant, \(G=6.67408\ 10^{-11}\),

The expression derived above for the energy of a neutron particle is, \(4\pi \rho_0 R_0 A\). It is no surprise that this expression gives the correct value of energy since it contains the displacement parameter, \(A\), which was derived from the expression for gravitational acceleration, as described in the chapter on gravitation. We have
\begin{align}
From\ gravitational\ theory,\ for\ a\ neutron\ A =&\ (\frac{G}{Hc_0}) M_n\nonumber \\
From\ particle\ energy\ theory\ E_n =&\ (4\pi \rho_0 R_0)\ A \nonumber \\
So\ that\ particle\ energy\ E_n =&\ (4\pi \rho_0 R_0)( \frac{G}{Hc_0}) M_n \nonumber
\end{align}
Where \(E_n\) is energy content of neutron particle, and \(M_n\) is neutron mass.

Since it is known that \(E_n = M_n c_0^2\), using the above values for Hubble constant, \(H\), Expansion velocity, \(c_0\), and Universal gravitational constant, \(G\), together with values of background Energy Density, \(\rho_0 = 7.48934\ 10^{-10}\), and Radius of the continuum, \(R_0 =1.1340097\ 10^{26}\), it is easily shown that, in numerical terms, \begin{align} c_0^2 = [4\pi \rho_0 R_0 \frac{G}{Hc_0}] \nonumber \end{align}

This implies that the expression can also be deduced via established relationships between these constants. It has already been shown in the chapter on gravitation that the universal gravitational constant, \(G\), is given by \begin{align} G = \frac{ H^2 c_0^2}{4 \pi \rho_0}\ \ so\ that\ \frac{4 \pi \rho_0 G}{H c_0} = H c_0\ \nonumber \end{align}
Since \(H=c_0/R_0\), this becomes \begin{align} \frac{4 \pi \rho_0 G}{H c_0} = \frac{c_0^2}{R_0}\ \ \ and\ hence\ \ \ \ \frac{4 \pi \rho_0 R_0 G}{H c_0} = c_0^2 \nonumber \end{align}

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