Number of Bits in the Universe - juhameskanen/abstract GitHub Wiki

Step 1: The Ehrenfest Bit-Flip Dynamics

We start with a system of $n$ bits. At step $0$, the system is in its ground state of zero entropy (all bits are $0$).

At each discrete raw tick (which we identify as a Planck time step), we choose one of the $n$ bits uniformly at random and flip it. This is a random walk on an $n$-dimensional hypercube.

Let $p(t)$ be the probability that any given individual bit is a $1$ at step $t$. The recurrence relation for this probability from step $t$ to $t+1$ is:

$$ p(t+1) = p(t)\left(1 - \frac{2}{n}\right) + \frac{1}{n} $$

Solving this linear recurrence with the initial condition $p(0) = 0$ gives:

$$ p(t) = \frac{1}{2}\left(1 - \left(1 - \frac{2}{n}\right)^t\right) $$

For $n$ much larger than $t$ (which we will assume and verify), we can use the standard exponential approximation $(1 - 2/n)^t \approx e^{-2t/n}$. At the end of inflation, where $t = T$:

$$ p(T) \approx \frac{1}{2}\left(1 - e^{-2T/n}\right) $$

Step 2: The Entropy of the System

The Shannon entropy of a single bit with probability $p$ of being $1$ is:

$$ S(p) = -p \log_2(p) - (1-p) \log_2(1-p) $$

Since $n$ is going to be astronomically larger than $T$, the ratio $\tau = T/n$ is incredibly small. This means $e^{-2T/n} \approx 1 - 2T/n$, which simplifies our bit probability to:

$$ p(T) \approx \frac{1}{2}\left(1 - \left(1 - \frac{2T}{n}\right)\right) = \frac{T}{n} $$

For an extremely small probability $p \approx T/n$, the second term of the Shannon entropy equation $(1-p)\log_2(1-p)$ is negligible. The entropy per bit is dominated by the first term:

$$ S(p) \approx -p \log_2(p) \approx -\frac{T}{n} \log_2\left(\frac{T}{n}\right) = \frac{T}{n}\left(\log_2(n) - \log_2(T)\right) $$

The total entropy of the entire $n$-bit system is:

$$ H_{\text{total}} = n \cdot S(p) \approx n \cdot \frac{T}{n}\left(\log_2(n) - \log_2(T)\right) = T\left(\log_2(n) - \log_2(T)\right) $$

Step 3: Solving for $n$

Find the value of $n$ where the total system entropy at time $T = 1.86 \times 10^8$ has reached $D = 6.19 \times 10^{60}$.

Setting $H_{\text{total}} = D$:

$$ T\left(\log_2(n) - \log_2(T)\right) = D $$

Divide both sides by $T$:

$$ \log_2(n) - \log_2(T) = \frac{D}{T} $$

$$ \log_2(n) = \log_2(T) + \frac{D}{T} $$

Now we plug in our values:

$T = 1.86 \times 10^8$
$D = 6.19 \times 10^{60}$

$$ \log_2(T) = \log_2(1.86 \times 10^8) \approx 27.47 $$

$$ \frac{D}{T} = \frac{6.19 \times 10^{60}}{1.86 \times 10^8} \approx 3.328 \times 10^{52} $$

Because $\frac{D}{T}$ is absolutely massive compared to $\log_2(T)$, we can comfortably write:

$$ \log_2(n) \approx 3.328 \times 10^{52} $$

To convert this to a base-10 exponent ($\log_{10}$):

$$ \log_{10}(n) = \log_2(n) \cdot \log_{10}(2) \approx 3.328 \times 10^{52} \cdot 0.30103 \approx \mathbf{1.002 \times 10^{52}} $$

The Verdict

The temporal resolution behaves as a strict Ehrenfest process.
The target physical scale $D$ represents the total entropy budget generated by the system ( $H_{\text{total}} = D$ ).

If these two premises hold, then the length of the bitstring $n$ required to reach that entropy state in that timeframe is an ultra-exponential number:

$$ \boxed {\LARGE n \approx 10^{1.002 \times 10^{52}}} $$

Note well: The value was calculated by factoring in the size and duration of the inflationary expansion era as predicted by the $\Lambda\text{CDM}$ model. Yet, no one was actually there to observe whether inflation occurred, or if it is simply a narrative we invented to reconcile General Relativity and Quantum Field Theory with the universe we see today. Ultimately, plugging these cosmic parameters into an information-theoretic framework most likely makes no sense.