Proof of Hyperbola
We have seen that, in the theory of varying elementary particle mass, atomic clock seconds are inversely proportional to mass:
$$
\frac{\Delta t}{\Delta t_0} = \frac{m_0}{m}
$$
Here $\Delta t=$ atomic clock unit of time, $\Delta t_0=$ the same at some reference time $\,t=0$ , $m=$ varying elementary
particle mass, $m_0=$ the same at some reference time $\,t=0$ . Apart from atomic clocks, there exists gravitational clocks.
Gravitational clocks are running at a different speed, namely squared the speed of an atomic clock:
$$
\frac{\Delta T}{\Delta T_0} = \left(\frac{m_0}{m}\right)^2
$$
Here $\Delta T=$ gravitational clock unit of time, $\Delta T_0=$ the same at some reference time $\,t=0$ . Atomic clocks and
gravitational clocks are synchronized at reference time. This means that : $\Delta T_0 = \Delta t_0$ . Consequently, while
replacing finite differences by infinitesimal differences:
$$
\frac{dT}{dt} \approx \frac{\Delta T}{\Delta t} = \frac{\Delta T}{\Delta T_0}\frac{\Delta t_0}{\Delta t} =
\frac{(m_0/m)^2}{m_0/m} = \frac{m_0}{m}
$$
Gravitational time itself can be recovered from its time units by counting and summing these time units from $1$ up to $N$ -
the clock is ticking, so to speak:
$$
T = \sum_k \Delta T_k = \frac{\Delta T_0}{\Delta m} \sum_{k=1}^N \left(\frac{m_0}{m_0 + k\cdot\Delta m}\right)^2 \cdot \Delta m
$$
Where it is noticed that $\,T=N\cdot\Delta T_0\,$ for $\,\Delta m = 0$ , i.e. when there is no varying mass; this is a familiar
result to all of us. But if there does exist a varying mass, then it's advantageous to approximate the sum by an integral.
Identify the varying mass as $\,\mu = m_0 + k\cdot\Delta m\, \to \, m = m_0 + N\cdot\Delta m$ . Then we have:
$$
T \approx \frac{\Delta T_0}{\Delta m} \int_{m_0}^m \left(\frac{m_0}{\mu}\right)^2 d\mu = \frac{\Delta T_0\cdot m_0^2}{\Delta m}
\int_{m_0}^m \frac{d\mu}{\mu^2} = \frac{\Delta T_0\cdot m_0^2}{\Delta m} \left[ - \frac{1}{\mu} \right]_{m_0}^m \\
= \frac{\Delta T_0\cdot m_0^2}{\Delta m} \left(\frac{1}{m_0}-\frac{1}{m}\right) = \frac{\Delta T_0\cdot m_0}{\Delta m}
\left[ 1 - \frac{m_0}{m} \right]
$$
Substitute the expression for $\,dT/dt\,$ herein and solve for $\,dt/dT$ :
$$
T \approx \frac{\Delta T_0\cdot m_0}{\Delta m} \left[ 1 - \frac{dT}{dt} \right]
\quad \Longrightarrow \quad \frac{T}{\Delta T_0\cdot m_0/\Delta m} = 1 - \frac{dT}{dt} \quad \Longrightarrow \\
\frac{dT}{dt} = 1 - \frac{T}{\Delta T_0\cdot m_0/\Delta m} \quad \Longrightarrow \quad
\frac{dt}{dT} = \frac{A}{A - T} \quad \mbox{with} \quad A = \frac{\Delta T_0}{\Delta m/m_0}
$$
This is quite similar to the expression that we have found before, having $\,T_0=0\,$ without loss of generality.
At reference time $\,T=t=0\,$ it is also the case that:
$$
\left.\frac{dt}{dT}\right|_{T=0} = 1 \quad \mbox{equivalent with} \quad \Delta T_0 = \Delta t_0
$$
The constant $A$ is the gravitational time unit divided by the percentage elementary particle mass increase. In practice,
though, it is a quantity of time that has to be determined empirically. In fact, it corresponds with the beginning of
gravitational time, since at $\,T=A\,$ atomic time is infinite and there is no time beyond that. The time units $\Delta T_0 = \Delta t_0$
are inversely proportional with mass, meaning that they are decreasing while the mass is increasing. So it is certain that creation time
$A = (\Delta T_0)/(\Delta m/m_0)$ must be negative when compared with today : $A < 0$ .
As for Light speed against time, there is no deviation from Setterfield's
theory by stating that The graph also represents the rate of ticking of the atomic clock against orbital time [ .. ]
The truth of the matter is, these are all the same curve.
Moreover it simply follows from the chain rule for differentiation in combination with the definition of the light speed in atomic
time units and gravitational time units - $\,ds=$ distance travelled - that the curves are indeed the same (though different from
the curves in Setterfield's theory):
$$
\frac{c_G}{c_0} = \frac{ds}{dT}/c_0 = \frac{ds}{dt}\frac{dt}{dT}/c_0 = c_0/c_0 \frac{dt}{dT} = \frac{dt}{dT}
\quad \Longrightarrow \quad c_G = c_0 \frac{A}{A-T}
$$