This can also be shown as a collection of manifold here.
The values of the scalar field at every location greater than the radius of the mass generating the gravitational field is:
\(
\begin{equation}\tag{1}
A_g = 1 + \frac{T_g – V_g}{E_0}
\end{equation}
\)
Performing calculations over the field in some instances is rather straight forward, for example the time rate difference between a position at sea level and the time rate at a position above sea level on a mountain. A not so straight forward calculation for example is evaluating the forces on an electron moving in a magnetic field over the scalar field. Because the unit distance and unit time are changing each infinitesimal step, an integration must find a way to represent this.
Consider a circular orbit (a fixed radius) in the gravitational field about a massive object. As observed from exemplar space the orbiting particle appears to maintain a pure Newtonian based orbit, starting at one point and returning to that point every revolution. To calculate the transformation from a non-exemplar space location to exemplar requires squaring the value of A. The reason is both distance and time (period of orbit) is transformed (distance multiplied by time). The transformation takes the form
\(
\begin{equation}\tag{2}
C=2 r \pi = 2\pi A r
\rightarrow 2\pi \left( 1 + \frac{G M}{2 r c^2} + \frac{G M}{r c^2} \right)^2r
\end{equation}
\)
Where the potential field is calculated from a gravitational orbit. Of interest about this transformation is the Circular orbit has a first order advance of
\(
\begin{equation}\tag{3}
advance = 2\pi \left( 1 + \frac{3 G M}{ c^2}\right)^2
\end{equation}
\)
Note the radial value r cancels out and what is left is a constant advance in length for all circular orbits regardless of radial distance from the central massive object. For orbits that are not circular but elliptical, the radial distance can be calculated as a dependency on the angle of rotation where r equals:
\(
\begin{equation} \tag{4}
r = \frac{a(1-e^2)}{1+e cos(\theta)}
\end{equation}
\)
Where a is the semi-major axis, e is the eccentricity, and $t\theta$ is the angle of rotation. Integrating the value for r over the angle of 2 Pi yields the advance. The full calculation is
\(
\begin{equation}\label{eqAdvance}\tag{5}
\int_{0}^{1}\int_{0}^{2\pi} \left(\left( 1 + \left( \frac{G M}{2 \left(\frac{a \left(1-e^2 \right)}{1+e cos(\theta)} \right) c^2}\right) + \left( \frac{G M}{\left(\frac{a \left(1-e^2 \right)}{1+e cos(\theta)} \right) c^2} \right) \right)^2 -1 \right) d\theta dP
\end{equation}
\)
The -1 is applied such that only the advance is calculated and not the entire elliptical circumference. If equation 5 is applied to the planet Mercury, it predicts an advance of 42.98 arc seconds per century. Charting the advance calculation for each planet, the predicted advance to observed advance is:
Of note, Figure 1 only takes into account the gravitational field produced by the sun. As the radial distance increases, other planet’s mass will have an impact, thus a plausible reason for some discrepancy in the chart.
Next -> Newtonian Mechanics Applied to the Hydrogen Atom Over the Scalar Field
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