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Milne's Formula

$ \def \hieruit {\quad \Longrightarrow \quad} \def \slechts {\quad \Longleftrightarrow \quad} \def \SP {\quad ; \quad} \def \MET {\quad \mbox{with} \quad} $ Quoting from What's the point of the Hoyle–Narlikar theory of gravity? (Physics Stack Exchange): Hoyle was really opposed to the big bang cosmological model, hence he really stuck with the quasi-static model all the way. [ .. ] Essentially, the advantage of the quasi-static model is that there is no "Why" left - you take the existence of the universe as a given. With a point of origin, like the Big Bang, you are then left asking why the Big Bang happened. [ .. ] in that model the universe did not 'happen': it has always existed. In short: The Big Bang Never Happened.

But how would this be possible? Haven't we defined just a minute ago that elementary particle rest mass is created at timestamp $T=A\,$? The only way out of this apparent paradox is to assume once again that orbital time and atomic time are different, so different that the universe has a beginning Alpha ($A$) in orbital time and it has always existed when measured in atomic time.
Two results from the previous, Time Dilation, are repeated for convenience: $$ \frac{dT}{dt} = \sqrt{\frac{m}{m_0}} \slechts \frac{dt}{dT} = \sqrt{\frac{m_0}{m}} $$ Combining this with (our version of) Narlikar's Law, we obtain (trivially assuming that $T \ge A$): $$ \frac{dt}{dT} = \sqrt{\frac{m_0}{m}} = \frac{T_0-A}{T-A} $$ Here $T=$ orbital time, $m=$ elementary particle rest mass, $m_0=$ reference mass (here and now), $T_0=$ gravitational reference time, which is the "nowadays" timestamp, $A=$ gravitational timestamp corresponding with a beginning (Alpha), at the time when rest mass is created out of nothing.
What we have now is an ordinary differential equation which can be solved easily: $$ t(T) = C\ln\left(\frac{T-A}{T_0-A}\right)+D $$ Where $A$ still is the time of creation. $C$ and $D$ are constants to be determined by setting the atomic increment equal to the orbital increment at the reference timestamp $(0)$: $$ \left.\frac{dt}{dT}\right|_{T=T_0} = 1 = \frac{C}{T_0-A} \quad \Longrightarrow \quad C = T_0-A \quad \Longrightarrow \\ t = (T_0-A)\ln\left(\frac{T-A}{T_0-A}\right) + D $$ Furthermore we must synchronize the clocks at the reference timestamp $(0)$ such that from $t=t_0$ it follows that $T=T_0$ and so $t_0=T_0$: $$ t_0 = (T_0-A)\ln\left(\frac{T_0-A}{T_0-A}\right) + D = D \quad \Longrightarrow \quad D = t_0 \quad \Longrightarrow \\ \large \boxed{\;t-t_0 = (T_0-A)\ln\left(\frac{T-A}{T_0-A}\right)\;} $$ A picture says more than a thousand words. It is clearly seen that orbital time $T$ has a beginning $A$ - Alpha, the moment of creation - while atomic time $t$ has no beginning, it extends to minus infinite. In addition, both timespans $T$ and $t$ have no ending in our theory. At last, it should be noticed that

  1. atomic clock ticks $\,dt\,$ are greater than orbital clock ticks $\,dT\,$ for $\,T < T_0\,$ and $\,t < t_0\,$ (the past)
  2. atomic clock ticks $\,dt\,$ are smaller than orbital clock ticks $\,dT\,$ for $\,T > T_0\,$ and $\,t > t_0\,$ (a future)

So far so good. But this is what we found on the internet. From Euclid to Eddington: A Study of Conceptions of the External World, by Edmund Taylor Whittaker. From the chapter 19. THE BEGINNING AND END OF THE WORLD:

Sober thinking reveals that professor Milne has discovered the same formula as we have, but with a little twist it seems: orbital time and atomic time are exactly the other way around. How can that happen? Well, the secret is in time kept. If clock ticks are smaller i.e. the clock runs faster, then the time measured between two events seems to last longer, because there are more of these shorter clock ticks between these two events. In general we can say that time measured is the inverse function of clock speed. Which could have explained the above "paradox".
Quite unfortunately, though, it doesn't. Digging further in A.E. Milne's theories reveals that his formula is indeed the other way around - when compared with ours - and it is penetrating all of his work on the subject:
  1. Kinematics, Dynamics, and the Scale of Time
  2. Kinematics, Dynamics, and the Scale of Time-II
  3. Kinematics, Dynamics, and the Scale of Time-III
  Milne   UAC
  kinematic     orbital  
  dynamical   atomic  

I really don't understand Milne's way of deriving things quite well - and that's an understatement. What to think about the following utterings in the first reference at the opening page 324, where Milne claims that the laws of dynamics and the Newtonian approximation to the law of gravitation [ .. ] have been deduced rationally [ .. ] No appeal was made in the derivations to any empirical laws of dynamics or gravitation, or even to the principle of relativity or to the principle of the constancy of the velocity of light. No empirical laws?! Okay .. whatever; other titbits of prose from the same reference are quite worthwhile to consider. But, before doing so, we must complete the relationship between Milne's formulation and ours.
Last but not least, for the sake of being able to employ a Nondimensionalization (NDM) technique eventually, the Formula shall be slightly rewritten, as follows: $$ t-t_0 = (T_0-A)\ln\left(\frac{T-A}{T_0-A}\right) \slechts \frac{t-T_0}{T_0-A} = \ln\left(1+\frac{T-T_0}{T_0-A}\right) \\ \slechts y = \ln(1+x) \MET x = \frac{T-T_0}{T_0-A} \SP y = \frac{t-T_0}{T_0-A} $$ Then the dimensionless relationship between $\,x\,$ and $\,y\,$ is like in the above picture, with $\,t_0=T_0=0\,$ and a dimensionless place of the Asymptote $\,=(A-T_0)/(T_0-A)=-1\,$ for $\,x=A\,$.

MilneUACNDM
  $\tau = t_0\log\frac{t}{t_0}+t_0$     $t-T_0=(T_0-A)\ln\left(\frac{T-A}{T_0-A}\right)$     $y = \ln(1+x)$  

From which it can easily be derived that:

  Milne   UACNDM
  $t_0$     $T_0-A$  
  $t$     $T-A$     $t_0(1+x)$  
  $\tau$     $t-A$     $t_0(1+y)$  
  $\log$     $\ln$  

Now we can quote, with our own notation and interpretation, starting at page 328.

The quotation in red will be considered once again in our Galaxy project, where it is conjectured that de Sitter is wrong here.