(real-analysis) (functions) (physics)

Weird problem with two types of parameterized clocks

Suppose that we have two sorts of clocks, called type A and type G. Both types are steered by some positive parameter, called $m$, in such a way that: $$ \mbox{type A :} \quad \frac{\Delta t}{\Delta t_0} = \frac{m_0}{m} \\ \mbox{type G :} \quad \frac{\Delta T}{\Delta T_0} = \left(\frac{m_0}{m}\right)^2 $$ Where $\Delta t$ is the period (say "second") of the clock type A and $\Delta T$ is the period (say "second") of the clock type G. Furthermore $t$ is the time as measured by clock type A and $T$ is the time as measured by clock type G. At some reference time $(0)$ the clocks are synchronized. This means that, at that timestamp: $$ \Delta t = \Delta T = \Delta t_0 = \Delta T_0 \quad \mbox{and} \quad t = T = t_0 = T_0 $$ If the parameter $m$ does not change, then the two clocks obviously will stay synchronized forever. But suppose that instead we have the wild hypothesis that the parameter $m$ does change, in fact it increases monotonically. Then I have a problem that I haven't been able to solve for weeks:

Is it possible to say something about the time as measured by a clock type A when compared with the time as measured by a clock type G ? More exactly, is it possible to formulate functions $A$ and $G$ such that $T = G(t)$ and $t = A(T)$ ?

I am interested in large time scales. In comparison with those all clock ticks may be considered as infinitesimally small; it is expected that the functions $A$ , $G$ are continuous and differentiable.

Concise physics motivation

Answer

Indeed the question is unclear. So the close votes it has gathered in the past are justified. But it still remains sort of a challenge to show why it is unclear what I am asking.
The gist of an answer is in the fact that it is undefined how the parameter $m$ is incrementing. Which is more or less obvious, certainly if you know the answer already :-)
Let us take two extreme cases for illustration purposes: first (1) like an arithmetic series and second (2) like a geometric series. Then we shall show that the relationships between the two time-scales $t$ and $T$ with (1) and (2) are so much different that no sensible generalization is to be expected. Thus leading to the answer that that nothing sensible can be said about function behavior $G(t)$ and $A(T)$ if no further information is supplied. And unfortunately: it is not supplied indeed.

1. Arithmetic sequence like

Start at $T=t=0$ with $m=0$ and define increments - not coincidentally - as follows: $$ m = k^2 \Delta m_0 \quad \mbox{with} \; k = 1,2,3, \cdots $$ Then we have for large $t$ and $T$ the following approximations - with help of known expressions for the sums $\sum 1/k^2$ and $\sum 1/k^4$ : $$ t \approx \Delta t_0 \cdot \sum_{k=1}^\infty \frac{m_0}{k^2 \Delta m_0} = \Delta t_0 \cdot \frac{m_0}{\Delta m_0} \frac{\pi^2}{6} \\ T \approx \Delta T_0 \cdot \sum_{k=1}^\infty \left(\frac{m_0}{k^2 \Delta m_0}\right)^2 = \Delta T_0 \left(\frac{m_0}{\Delta m_0}\right)^2 \frac{\pi^4}{90} \\ $$ Solve $\,m_0/\Delta m_0\,$ from the first equation and substitute into the second one: $$ \frac{m_0}{\Delta m_0} \approx \frac{6}{\pi^2}\frac{t}{\Delta t_0} \quad \Longrightarrow \quad T \approx \Delta T_0 \frac{36}{\pi^4}\left(\frac{t}{\Delta t_0}\right)^2 \frac{\pi^4}{90} \\ \Longrightarrow \quad T \approx \frac{2\,t^2}{5\,\Delta t_0} $$ Therefore $T$ is quadratic in $t$ : $G(t) \sim t^2$ .

2. Geometric sequence like

Start at $T=t=0$ with $m=0$ and define increments $m_k > m_{k-1}$ - not coincidentally - as follows: $$ \frac{m_0}{m_1} = \frac{m_1}{m_2} = \frac{m_2}{m_3} = \cdots \quad \Longrightarrow \quad \frac{m_0}{m_k} = \left(\frac{m_0}{m_1}\right)^k $$ Then we have for large $t$ and $T$ the following approximations - with help of known expressions for sums of the form $\sum r^k$ : $$ t \approx \Delta t_0 \cdot \sum_{k=1}^\infty \left(\frac{m_0}{m_1}\right)^k = \frac{\Delta t_0}{1 - m_0/m_1} \\ T \approx \Delta T_0 \cdot \sum_{k=1}^\infty \left(\frac{m_0}{m_1}\right)^{2k} = \frac{\Delta T_0}{1 - (m_0/m_1)^2} $$ Solve $\,m_0/m_1\,$ from the first equation and substitute into the second one: $$ \frac{m_0}{m_1} \approx 1 - \frac{\Delta t_0}{t} \quad \Longrightarrow \quad T \approx \frac{\Delta T_0}{1 - (1 - \Delta t_0/t)^2} = \frac{\Delta T_0}{2\Delta T_0/t + (\Delta t_0/t)^2} \\ \Longrightarrow \quad T \approx \frac{t^2}{2\,t+\Delta t_0} \approx \frac{1}{2} t $$ Therefore $T$ is linear in $t$ : $G(t) \sim t$ .
But if a function can be linear as well as quadratic, then it can be virtually anything, dependent on how the parameter $m$ is incrementing. This completes the claim that the answer is unclear as well. Philisophical remark. Since everybody is trusting his own clock, it can not be observed that the sum of the "seconds" converges to a finite time interval, in all of the above cases. The only thing that can be observed eventually is that the two clocks are running out of sync.