c·t = CONST, THAT CLICKS EPISODE 24 ／ The rate capacity grows is not a communication speed
Bandwidth × age = memory — exactly one pass
How many bits per second
cross the horizon?
Episode 17 said "there was no way to send the 20 KB". Here is the bandwidth.
And bandwidth turns out not to be the bottleneck at all.
Episode 17 counted "the agreement needed 20 KB; the problem was that there was no channel". So how wide is the channel? Episode 1 gave \(dN/dt=1.36\times10^{105}\) bit/s, but that is the rate at which capacity grows, not a communication speed. Today we compute the bandwidth itself — and between the two numbers there falls out a rather clean identity.
01Two different "bits per second"
| Quantity | Meaning | Value |
|---|---|---|
| \(dN/dt\) | rate at which capacity grows (Episode 1) | \(1.36\times10^{105}\) bit/s |
| \(C\) | channel capacity (today) | \(6.79\times10^{104}\) bit/s |
The bandwidth comes from dividing the Bekenstein bound by the time a signal takes to cross. Information in a system of radius \(R\) is at most \(S\le2\pi ER/\hbar c\), and the crossing time is \(R/c\), so —
with the total energy inside the horizon \(E=7.90\times10^{69}\) J
$$C=6.79\times10^{104}\ \mathrm{bit/s}$$02The heart — bandwidth × age = memory
The ratio of the two numbers is exactly 2.000000. Not a coincidence.
substituting \(E=c^4R/2G\) and \(\ell_P^2=\hbar G/c^3\)
$$C=\frac{2\pi}{\hbar\ln2}\cdot\frac{c^4R}{2G}=\frac{\pi cR}{\ell_P^2\ln2}$$while \(N=\pi R^2/(\ell_P^2\ln2)\) with \(R=ct\), so
$$C\cdot t=\frac{\pi cR}{\ell_P^2\ln2}\cdot\frac{R}{c}=\frac{\pi R^2}{\ell_P^2\ln2}=N$$The thing this episode most wants to say
$$\boxed{\ C\cdot t=N\ }$$The universe has exactly enough bandwidth to move its entire memory once per Hubble time.
── No more and no less: exactly once.
And since \(N\propto t^2\), \(dN/dt=2N/t=2C\) — the factor of two between Episode 1's number and today's was two faces of the same identity.
| Reading | Formula | Meaning |
|---|---|---|
| as bandwidth | \(C\cdot t=N\) | one full pass of memory per Hubble time |
| as capacity growth | \(dN/dt=2C\) | memory grows at twice the rate it can be moved |
By Episode 19's classification this is an identity — 0 bits of surprise. It follows automatically from \(E=c^4R/2G\) (Dirac's large numbers) and holography. It can still be used to read the design of the universe-as-computer.
03Episode 17's 20 KB could have been sent instantly
With bandwidth in hand we can settle Episode 17's homework — how long would the "20 KB to be agreed on" have taken to send?
| Epoch | Bandwidth \(C\) | Time to send 20 KB |
|---|---|---|
| Nucleosynthesis (\(t=1\) s) | \(1.56\times10^{87}\) bit/s | \(1.0\times10^{-82}\) s |
| Recombination (380,000 yr) | \(1.87\times10^{100}\) bit/s | \(8.6\times10^{-96}\) s |
| Today | \(6.79\times10^{104}\) bit/s | \(2.4\times10^{-100}\) s |
Conclusion of §03
Given a channel, the horizon problem's 20 KB could have been sent in \(10^{-96}\) seconds.
The problem was never bandwidth but the absence of a channel — Episode 17's conclusion, confirmed numerically.
Figure: bandwidth \(C\) by epoch (slope 1) against the memory \(N\) at that time (slope 2). Since \(C\cdot t=N\), the two lines are always separated by exactly \(t\). Move the slider to read off how long 20 KB takes to send.
04Bandwidth per particle
| System | Energy | Bandwidth in principle |
|---|---|---|
| One CMB photon | \(2.35\times10^{-4}\) eV | \(3.2\times10^{12}\) bit/s |
| One proton | 938 MeV | \(1.3\times10^{25}\) bit/s |
| 1 kg of matter | \(9.0\times10^{16}\) J | \(7.7\times10^{51}\) bit/s |
A single CMB photon could in principle carry three trillion bits per second. What it actually carries is a few bits (temperature and polarisation) — the same picture again of "capability entirely unused".
total world internet traffic (roughly)
$$1.3\times10^{15}\ \mathrm{bit/s}$$against the in-principle bandwidth of 1 kg of matter
$$1.7\times10^{-37}\qquad(\text{against the horizon, }1.9\times10^{-90})$$05The reveal — the underuse is not a shortfall of capability
Episode 6: "only \(10^{-18}\) of the memory is used". Episode 22: "95% of the operational budget goes to components in which nothing happens". Today: "bandwidth is one memory pass". Three routes to the same conclusion — the universe as a computer is doing overwhelmingly little relative to its specification.
① \(C=2\pi E/(\hbar\ln2)\) is the Bekenstein bound divided by a crossing time. It is the same kind of quantity as the Bremermann limit, but the coefficient depends on how the derivation is set up (the shape of the boundary, the choice of \(R\)) — factors of order \(\pi\) move. It differs from Episode 1's Margolus–Levitin limit (\(2E/\pi\hbar\)) by \(\pi^2\), which is also a matter of convention.
② \(C\cdot t=N\) is an identity (0 bits of surprise by Episode 19). It follows automatically from the Dirac large-number identity \(E=c^4R/2G\) and holography \(N\propto R^2\). It is not a physical claim that the universe "provides exactly the right bandwidth".
③ The send times in §03 divide by bandwidth alone. In reality the signal also needs time to cross the distance (\(R/c\) — 380,000 years at recombination). The calculation is there to show that bandwidth is not the bottleneck, not to say that 20 KB arrives in \(10^{-96}\) s.
④ "Three trillion bits per second per CMB photon" is likewise an in-principle bound. A real photon carries its frequency, polarisation and direction of arrival — a few bits.
⑤ The \(1.3\times10^{15}\) bit/s of internet traffic is an order-of-magnitude marker, moving by factors of a few with what counts as traffic.
Exercises (solvable with this episode's formulas alone)
- State the difference between \(dN/dt\) and \(C\).
Show the answer
\(dN/dt\) is the rate at which capacity grows (how much new writable space appears); \(C\) is the channel capacity (how much existing information can be moved). Entirely different quantities, joined by the identity \(dN/dt=2C\). - Show that \(C\cdot t=N\).
Show the answer
Substituting \(E=c^4R/2G\) and \(\ell_P^2=\hbar G/c^3\) into \(C=2\pi E/(\hbar\ln2)\) gives \(C=\pi cR/(\ell_P^2\ln2)\). Multiplying by \(t=R/c\) gives \(\pi R^2/(\ell_P^2\ln2)=N\). One full memory pass per Hubble time. - How long does 20 KB take to send at recombination?
Show the answer
\(C=1.87\times10^{100}\) bit/s, so \(1.6\times10^5/1.87\times10^{100}=8.6\times10^{-96}\) s. Bandwidth is not remotely the bottleneck — confirming Episode 17's "the problem is the existence of a channel". - What is one CMB photon's in-principle bandwidth?
Show the answer
\(C=2\pi E/(\hbar\ln2)\) with \(E=2.35\times10^{-4}\) eV \(=3.76\times10^{-23}\) J gives \(3.2\times10^{12}\) bit/s. What it actually carries is a few bits. - (Harder) What is the bottleneck of the universe-as-computer?
Show the answer
The wiring (causal structure). Only \(10^{-18}\) of the capacity is used (Episode 6), 95% of the operational budget goes to components in which nothing happens (Episode 22), and bandwidth provides one memory pass (today) — performance is everywhere in surplus. What limits it is that with no causal connection not one bit crosses however much bandwidth there is (Episode 17). And the wiring is set by the expansion law.
Summary — bandwidth × age = memory
We distinguished two "bits per second" — \(dN/dt=1.36\times10^{105}\) bit/s is the rate at which capacity grows, while \(C=2\pi E/(\hbar\ln2)=6.79\times10^{104}\) bit/s is the channel capacity, obtained by dividing the Bekenstein bound by a crossing time.
The ratio is exactly 2.000000, and it is an identity — \(E=c^4R/2G\) and holography give \(C=\pi cR/(\ell_P^2\ln2)\), hence \(C\cdot t=N\). The universe has exactly enough bandwidth to move its entire memory once per Hubble time — no more, no less. The factor of two against Episode 1 is just the flip side of \(N\propto t^2\).
That settles Episode 17's homework. At recombination's bandwidth, the horizon problem's 20 KB could be sent in \(10^{-96}\) seconds. Bandwidth is not remotely the bottleneck — the problem was not the channel's width but its existence.
And three routes converge: only \(10^{-18}\) of the capacity is used (Episode 6), 95% of the operational budget goes to components in which nothing happens (Episode 22), and bandwidth provides a full memory pass (today). "Unused" is not a shortfall of capability. The bottleneck of the universe-as-computer is neither bandwidth nor capacity but wiring — the causal structure, which the expansion law determines.
This document is Episode 24 of "c·t = const, That Clicks", written for physics-minded high-school and university readers. The Bekenstein bound \(S\le2\pi ER/\hbar c\) is standard, and dividing it by a crossing time \(R/c\) to obtain a channel capacity \(C=2\pi E/(\hbar\ln2)\) is the same kind of construction as the Bremermann limit, but the coefficient depends on how the derivation is set up (the shape of the boundary, the choice of \(R\)) — the \(\pi^2\) difference from Episode 1's Margolus–Levitin limit (\(2E/\pi\hbar\)) is likewise conventional. The values \(C=6.79\times10^{104}\) bit/s, \(C=\pi cR/(\ell_P^2\ln2)\), and the identities \(C\cdot t=N\) and \(dN/dt=2C\) are derived here (kenshou/calc28.py). These are identities — they follow automatically from the Dirac large-number identity \(E=c^4R/2G\) and holography \(N\propto R^2\) — and are not a physical claim that the universe "provides exactly the right bandwidth" (0 bits of surprise by Episode 19's classification). The send times in §03 divide by bandwidth alone; the time for a signal to cross the distance (380,000 years at recombination) is additional — the calculation is there to show bandwidth is not the bottleneck. The per-particle bandwidths are likewise in-principle bounds; a real CMB photon carries a few bits (frequency, polarisation, direction). The internet's \(1.3\times10^{15}\) bit/s is an order-of-magnitude marker. Linear expansion (\(c\cdot t=\)const, \(R_h=ct\)) is a minority model under examination, and \(R_H=ct\) is its convention (in \(\Lambda\)CDM, \(R_H=c/H_0\) differs from the particle horizon). The academic standard remains the \(\Lambda\)CDM model including inflation. ── To make a PDF, use your browser's Print dialogue (sliders freeze and answers are hidden in the print version).