De Sitter - Ehrenfest Mapping

Given $n$ bits, $2^n$ configurations, sorted in increasing entropy order describes typical relaxed entropy driven process rushing from zero entropy towards full equilibrium.

Entropy $S(\tau) = H_2(p(\tau))$ is bounded — it goes from $0$ to $1$ bit as $\tau \to \infty$. Therefore, entropy itself cannot be made to match an unbounded scale factor of De Sitter by any relabeling of time. Bounded and unbounded functions just can't be reconciled that way, no matter what substitution one uses.

But $p(\tau)$ itself has a different, exact algebraic structure in it. Since $1-2p(\tau) = e^{-2\tau}$ exactly, its reciprocal is

$$ \frac{1}{1-2p(\tau)} = e^{2\tau} $$

which is unbounded, growing exponentially in $\tau$. If we identify $\tau = \frac{H}{2}t$ (a linear, not logarithmic, relationship between flip-time and physical time), this becomes

$$ \frac{1}{1-2p(\tau)} = e^{Ht} = a(t) $$

exactly — an exact algebraic match to the de Sitter scale factor.

We defined $\tau = t_\text{raw}/n$ specifically to remove the size-dependence from the mean-field ODE. That rescaling is why $p(\tau)$ has that clean, $n$-independent closed form — but it also means $\tau$ reaching a modest value (like 30) doesn't mean only 30 actual bit-flip events have happened. It means $30n$ have.

So when we identify $\tau = \frac{H}{2}t_\text{phys}$ (needed for the exact match to $a(t)=e^{Ht}$) and also $\tau = t_\text{raw}/n$ (its definition), combining these forces:

$$ H = \frac{2}{n} $$

Conclusions

Entropy $S(\tau)$ and this reciprocal quantity are both functions of $p(\tau)$, but they behave completely differently at large $\tau$ (one saturates, one blows up) — which one is "the scale factor"?

Why would $1/(1-2p)$ specifically represent physical size?

If we can tie $1/(1-2p)$ to the actual decoding scheme, this would be a genuinely strong result — an exact, not approximate, correspondence, with a physically clean bonus: as $\tau\to\infty$ (equilibrium/heat death), $a\to\infty$, so "the universe expands forever, faster as it approaches maximum entropy" falls right out, and it never actually reaches equilibrium in finite $\tau$ — matching the usual heat-death-as-asymptote picture.

This is a linear $\tau \leftrightarrow t$ relationship, which is a different animal from the exponential $t_\text{flips}\sim\ln(t_\text{physical})$ ansatz have proposed in Paper IV wiki page.

$$ H_\text{model} = \frac{2}{n} \approx 1.15\times10^{-122} \quad (\text{in inverse-Planck-time units}) $$

$$ H_\text{real, inflation} \approx 3.23\times10^{-7} \quad (\text{same units, from } N{=}60 \text{ e-folds over } \Delta t \approx 10^{-35}\text{s}) $$

$$ \frac{H_\text{model}}{H_\text{real}} \approx 3.6\times10^{-116} $$

That's the same ~115-order-of-magnitude gap from the wiki page, arrived at independently. Which is actually the right outcome to expect: this is the same underlying data (same $n$, same $\Delta t$), just run through a different piece of algebra, so of course it reproduces the same tension rather than dissolving it. A single dataset can't have two different answers depending on which formula you use to look at it — if it had come out consistent, that would have been the surprising (and suspicious) result. But this shows we are on the right track.

$1/(1-2p(\tau))$ gives an exact, non-asymptotic match to $a(t)=e^{Ht}$ under a clean linear time identification.

Anyway, the solution collapses two previously separate-looking problems (the shape mismatch in Paper III's discussion, and the tick-budget mismatch in Paper IV Section 4) into one single quantity, $H = 2/n$.

Instead of two different symptoms, we have one clean, falsifiable relation to interrogate: why does the model predict such a tiny effective $H$, and what would have to change (parallelism, effective smaller $n$ during the inflationary regime specifically, a different tick-to-Planck-time mapping) to make $2/n_\text{eff}$ come out at $\sim10^{-7}$ instead of $\sim10^{-122}$?*

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