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3.5.3 Particle Volume and Energy Density

Spherical symmetry of displacement allows a simple visualization of particle structure, which we imagine to consist of concentric spherical shells, each displaced into the fourth dimension by the displacement function, \(N=-A/r\). The truncation of displacement and energy density at the edge of the core zone imposes spherical curvature on the particle structure. The curvature, \(2/r\), acts on each of the spherical shells. Hence, the energy density is highest at the edge of the core zone, \(r=r_0\), and decreases as radius increases.

In deriving an equation which represents energy density and gradient within a particle, it would be necessary to have a formal, four dimensional mathematical framework. However, since this is not yet available, the following three dimensional explanation using displacement gradient, suffices to provide the correct relationship. \begin{align} energy\ density\ = \rho_0 2 \frac{R_0}{r} \nonumber \end{align} To find the increment of energy in the spherical shell at radius, r, it is necessary to multiply energy density by the incremental volume of the shell, which is \(4 \pi r^2 \partial r\). \begin{align} Energy\ in\ shell\ at\ radius\ r\ = \rho_0 2 \frac{R_0}{r}\ 4 \pi r^2 \partial r = 4 \pi \rho_0 R_0\ 2 r \partial r \nonumber \end{align}

To find the total energy of all the spherical shells up to a radius, \(r\), it is necessary to integrate this expression between limits, \(r=r_0\) and \(r=r\), as follows. \begin{align}
Energy\ in\ structure\ up\ to\ radius\ r, \nonumber \\
\int_{r_0}^{r} 4 \pi \rho_0 R_0\ 2 r\ \partial r = 4 \pi \rho_0 R_0\ (r^2-r_0^2) \nonumber \end{align}

To take into account the curvature due to displacement it is necessary to multiply this result by displacement gradient, \(A/r^2\).

\begin{align} particle\ energy\ up\ to\ radius\ r =&\ 4\pi\ \rho_0\ R_0\ \frac{A}{r^2} (r^2- r_0^2) \nonumber \\
=&\ 4\pi\ \rho_0\ R_0\ A (1- \frac{r_0^2}{r^2}) \nonumber
\end{align}

This relationship is valid for \(r>r_0\), that is, outside the core zone. At the edge of the core zone, \(r=r_0\), and the term, \((1- \frac{r_0^2}{r^2})\) is zero.

Outside the core zone where, \(r>r_0\), the term, \((1- \frac{r_0^2}{r^2})\), is finite and positive, and approaches one as radius approaches infinity, so that the total energy of the particle becomes \(4 \pi \rho_0 R_0 A\).

For a neutron, which has a mass, \(M=1.67493\ 10^{-27}\) Kg, will have a displacement constant, \(A=1.4104729\ 10^{-28}\), square meters. Putting this value into the expression for total particle energy, \( 4\pi \rho_0 R_0 A\), gives, \(1.5053411\ 10^{-10}\) Joules, which is correct to the fifth decimal place.

Finally, it is important to note that this expression for Neutron energy is the result of a principled analysis of its structure, and the fact that this gives the correct value may be accepted as confirmation that this visualization of its structure is correct. The constants included in this expression are also the result of principled analysis and the relationships between them suggest that the concept of displacement is correct.

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