Recall \(\phi\), the scalar value transformation is
\begin{equation}
\phi^{-1} = 1 + \frac{T_g + V_g + V_t}{E_0}
\end{equation}
In the first investigation, the \(T_g\) value is zero since the clock’s time interval is not experiencing acceleration. Second, if only considering time, then the change in space is zero. Thus, the transformation is
\begin{equation}
\phi^{-1} = 1 + \frac{V_t}{E_0} = 1 + \frac{G M}{r c^2}
\end{equation}
A clock’s interval \(\delta t_e\) in exemplar space will transform as an interval \(\delta t_p\) in privo space via:
\begin{equation}
t_p = \phi^{-1} t_e = \left(1+ \frac{G M}{r c^2}\right) t_e
\end{equation}
Since \(\phi^{-1}\) must be equal to one or greater, the time interval \(t_e\) is observed as longer (ticks slower) in privo space where \(t_p\) is observed. So, how does this translate to a physical world example?
Static Transformation
Let us return to the clock’s time interval measured in a valley, compared to a clock’s time interval measured on a mountain. For this example, the rotation of the Earth is not considered; thus, only the static (no acceleration) transformation will be used. If the mountain is 1,000 meters above sea level, and the valley is at sea level, then in the valley, the distance the clock is from the center of the Earth is \(r_v=R_{earth}\), and for the clock on the mountain, it is \(r_m=R_{earth} + 1000m\). Placing this into the transformation, one has:
\begin{equation}
\begin{split}
\phi_v^{-1} = \left(1+\frac{G M_{earth}}{r_v c^2} \right) \
\phi_m^{-1} = \left(1+\frac{G M_{earth}}{r_m c^2} \right)
\end{split}
\end{equation}
Next, the time intervals are to be transferred to exemplar space, making all units of measurement equal. The time interval observed does not magically change. They both observe the same passage of time when observing the same clock. Thus, their observances are equal and in equal units when transformed to exemplar space. Therefore, one can set the observed intervals equal and solve the differential in time dilation between the two locations.
\begin{equation}
t_{exemplar} = \phi_m t_m \end{equation}
\begin{equation}t_{exemplar} = \phi_v t_v \end{equation}
\begin{equation}t_{v} = \left(\frac{\phi_m}{\phi_v}\right) t_m \end{equation}
\begin{equation}t_{m} = \left(\frac{\phi_v}{\phi_m}\right) t_v
\end{equation}
Where \(t_v\) and \(t_m\) are the time intervals measured in the valley and the mountain, respectively. By using the mass of the Earth as \(5.9736e24kg\) and the radius of the Earth as \(6378100m\) and setting the two observances equal in exemplar space, the differential time dilation values are:
\begin{equation}
\phi_v = 0.99999999930465150891 \end{equation}
\begin{equation}\phi_m = 0.99999999930476051245 \end{equation}
\begin{equation}\left(\frac{\phi_m}{\phi_v}\right) = 1.00000000000010900354 \end{equation}
\begin{equation}\left(\frac{\phi_v}{\phi_m}\right) = 0.99999999999989099646 \end{equation}
\begin{equation}t_v = (1.00000000000010900354) t_m \end{equation}
\begin{equation}t_m = (0.99999999999989099646) t_v
\end{equation}
Thus, the clock in the valley runs slower than the clock on the mountain by a scaled value of (\(1.00000000000010900354\)). The value of \(\phi\) is unitless, thus a true scalar value. If the interval were seconds, for every 1 second as observed in the valley, the mountain would have \newline \(1.00000000000010900354\) seconds pass; thus, the mountain’s clock is slightly faster than the valley’s clock. This aligns with GR in that the deeper in a gravitational potential well a clock is, the slower the clock ticks compared to one an infinite distance from the gravitational potential well. Note, this NUVO transformation did not consider the rotation of the Earth and thus did not include any local effects (the \( \frac{v^2}{2 c^2}\) part of the transformation).
GPS Time Dilation
Many people often use the GPS as a “proof” of General Relativity and Special Relativity . As previously discussed, this author strongly disagrees with any use of SR for calculating values in a non-inertial frame; it is a gross misuse and misunderstanding of SR to do so. Any valid theory should be able to account for all-time dilation due to the acceleration (gravitational field or acceleration through the field) of a system. To calculate the time dilation of the GPS (compared to a clock on Earth), one follows the same method as the clock at sea level and the clock on the mountain. The earth’s rotation will not be added to the equations for this calculation as it adds no real contribution. The Earth’s surface rotates at about 460 meters per second compared to lights 299,792,458 meters per second, making the \(\frac{v^2}{2 c^2}\) value nearly trivial. The $\phi$ value for the Earth is: (Recall the radial distance to the satellite is constant)
\begin{equation}
\phi_E=\left(1+\frac{G M_E}{R_E c^2}\right)^{-1}
\end{equation}
\(\Phi\) for the satellite is:
\begin{equation}
\phi_S=\left(1+ \frac{G M_E}{2 R_S c^2} + \frac{G M_E}{R_S c^2} \right)^{-1}
\end{equation}
Where the middle term of the RHS is due to the instantaneous velocity experienced by the satellite. A static transformation from the satellite’s position (accounting for all accelerations, both global – GR and local – orbital acceleration) to the Earth’s surface position, one needs to transform both to Exemplar space. The inverse of \(\phi\) is used to transform to Exemplar space.
\begin{equation}
\phi_{earth}^{-1} = 1.00000000069534849157 \end{equation}
\begin{equation}\phi_{satellite}^{-1} = 1.00000000025045205937 \end{equation}
\begin{equation}t_{earth} = \left( \frac{ \phi_{satellite}^{-1}}{\phi_{earth}^{-1}} \right) \delta t \end{equation}
\begin{equation}t_{satellite} = \left( \frac{ \phi_{earth}^{-1}}{\phi_{satellite}^{-1}} \right) \delta t \end{equation}
If one is at sea level on the Earth, they only know the value of their time interval. For this calculation, on Earth at sea level, the time interval is nanoseconds in a day, which is \(86400000000000 ns\). Thus to know the dilation between the two locations, equate (\(t_{satellite}\)) and subtract the number of nanoseconds at sea level.
\begin{equation}
\begin{split}
t_{difference} = \left(t_{satellite}\right) t_{ns-in-a-day} – t_{ns-in-a-day} \
t_{difference} = \left( \frac{ \phi_{earth}^{-1}}{\phi_{satellite}^{-1}} \right) t_{ns-in-a-day} – t_{ns-in-a-day} \
t_{difference} = 38439 ns
\end{split}
\end{equation}
Thus the satellite gains approximately 38439 ns per day due to time dilation effects of gravity and acceleration. This corresponds to known measured effects and calculations when both GR and SR are used to calculate the dilation.