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 $\,t=t_0$ , $m=$ varying elementary
particle mass, $m_0=$ the same at that reference time.
Apart from atomic clocks, there exists gravitational clocks.
Gravitational clocks are running at a different speed, namely inversely proportional to mass squared:
$$
\frac{\Delta T}{\Delta T_0} = \left(\frac{m_0}{m}\right)^2
$$
Here $\Delta T=$ gravitational clock unit of time, $\Delta T_0=$ the same at our reference time $\,T=T_0$ . Which is at the
same time the atomic reference time $\, T_0=t_0$ , because the two types of clocks are being synchronized at that time.
Let analysis (calculus) start. Replace finite differences by infinitesimal differences:
$$
\frac{dt}{dm} = \frac{dt_0}{dm}\frac{m_0}{m} \\
\frac{dT}{dm} = \frac{dT_0}{dm}\left(\frac{m_0}{m}\right)^2
$$
Define $b$ and $B$ as the following constants:
$$
b = \frac{dt_0}{d(m/m_0)} \quad ; \quad B = \frac{dT_0}{d(m/m_0)}
$$
The meaning of these being how the time units are changing with a unit (elementary particle) mass change at the reference time stamp $\,t_0=T_0$ .
The reverse quantities are easier to understand: how the unit (elementary particle) mass is changing with atomic time and gravitational time
respectively, at the reference time stamp.
$$
\frac{d(m/m_0)}{dt_0} = \frac{1}{b} \quad ; \quad \frac{d(m/m_0)}{dT_0} = \frac{1}{B}
$$
Say that the time units are one split of a second. then there is an incredible small change of mass within that split of a second.
This means that $1/b$ and $1/B$ are very small positive numbers. Hence both $b$ and $B$ themselves must be very large positive numbers :
$b \gg 0\,$ and $B \gg 0$ .
Integrating the first equation gives:
$$
\frac{dt}{dm} = \frac{b}{m} \quad \Longrightarrow \quad t-t_0 = b\,\ln\left(\frac{m}{m_0}\right)
\\ \Longrightarrow \quad m = m_0\,e^{(t-t_0)/b}
$$
The (elementary particle) mass is slowly but exponentially increasing with atomic time. Integrating the second equation gives:
$$
\frac{dT}{dm} = \frac{B\,m_0}{m^2} \quad \Longrightarrow \quad
T = B\,m_0\left(-\frac{1}{m}\right) + C \\ \Longrightarrow \quad C = T_0 + B\,m_0\frac{1}{m_0}
\quad \Longrightarrow \quad T-T_0 = B\left(1-\frac{m_0}{m}\right)
$$
Solve $(m_0/m)$ from the second equation:
$$
\frac{m_0}{m} = 1-\frac{T-T_0}{B} \quad \Longrightarrow \quad \frac{m}{m_0} = \frac{B}{B-(T-T_0)}
$$
The (elementary particle) mass is hyperbolically increasing with gravitational time. Substitute into the first equation:
$$
\Longrightarrow \quad t-t_0 = b\,\ln\left(\frac{B}{B-(T-T_0)}\right) = - b\,\ln\left(\frac{B-(T-T_0)}{B}\right)
$$
The derivative of this function is:
$$
\frac{dt}{dT} = b \cdot \frac{1}{B-(T-T_0)} \quad \Longrightarrow \quad \left.\frac{dt}{dT}\right|_{T=T_0} = \frac{b}{B} = 1
\quad \Longrightarrow \quad b = B
$$
We know that the latter must be the case because the atomic clock and the gravitational clock are synchronized at reference time (0).
This applies not only to the time stamps $\,t_0=T_0\,$ but it applies to the time units (our "split seconds") as well : $dt_0 = dT_0$ .
Right ? At last, put $\,B = b = A-T_0$ , then:
$$
t-t_0 = (T_0-A)\ln\left(\frac{T-A}{T_0-A}\right) \\
\\ \Longrightarrow \quad \frac{dt}{dT} = \frac{T_0-A}{T-A} = \frac{m}{m_0}
$$
This proves that the rate of change of atomic time with respect to gravitational time is according to a hyperbola, with an asymptote
at a time $\,A \gg T_0$ . This is contrary to expectation, which is $\,A \ll T_0$ .
So we have a little problem, because there is an end-time $A$ but no creation time at all.
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) and, according to the above, increasing instead of decreasing ipse est (i.e.) the wrong
way around:
$$
\frac{c_G}{c_0} = \frac{ds}{dT}/c_0 = \frac{ds}{dt}\frac{dt}{dT}/c_0 = c_0/c_0 \frac{dt}{dT} = \frac{dt}{dT}
\quad \Longrightarrow \quad c_G = c_0 \frac{T_0-A}{T-A}
$$