c·t = CONST, THAT CLICKS EPISODE 18 ／ Holography read in the language of addressing
The volume one bit is responsible for has a side about the size of a proton
There are not enough
address lines
\(5.27\times10^{182}\) spatial cells, \(2.96\times10^{122}\) bits writable.
Only \(10^{-61}\) of the cells can be addressed — and the gap keeps widening.
Episode 17 was communication. Today: addressing. The universe has \((R_H/\ell_P)^3=5.27\times10^{182}\) spatial cells but can write only \(2.96\times10^{122}\) bits. Far from one bit per cell, there is \(10^{-61}\) of one. Read holography not as "written on the area" but as "there are not enough addresses", and a curiously specific length falls out along the way — the volume one bit is responsible for has a side about the size of a proton.
01Counting three numbers
today's universe in Planck lengths
$$\frac{R_H}{\ell_P}=8.075\times10^{60}$$spatial cells (Planck volumes)
$$\left(\frac{R_H}{\ell_P}\right)^3=5.27\times10^{182}$$four-volume cells (Planck spacetime points)
$$\left(\frac{ct_0}{\ell_P}\right)^4=4.25\times10^{243}$$bits writable (Episode 1)
$$N=\frac{\pi}{\ln2}\left(\frac{R_H}{\ell_P}\right)^2=2.96\times10^{122}$$The exponents are \(3\), \(4\) and \(2\). Only the bits go as a square — that is the whole of holography.
02The ratio is a clean identity
Conclusion of §02
$$\boxed{\ \frac{\text{bits writable}}{\text{spatial cells}}=\frac{\pi/\ln2}{R_H/\ell_P}\ }$$Holography means "only \(10^{-61}\) of the spatial cells can be given an address".
And since the ratio goes as \(1/R_H\), the shortage worsens as the universe grows.
Conversely, it was better in the past. Solving \(N=(R_H/\ell_P)^3\) gives \(R_H/\ell_P=\pi/\ln2=4.53\). Addresses sufficed only while the universe was smaller than 4.5 Planck lengths.
03The heart — the volume one bit is responsible for
as a volume
$$7.52\times10^{-45}\ \mathrm{m^3}\qquad\Longrightarrow\qquad \text{side}\ \ell_{\rm bit}=1.96\times10^{-15}\ \mathrm{m}$$The thing this episode most wants to say
The volume one bit is responsible for is a cube of side 1.96 femtometres.
── The size of a proton (charge radius 0.84 fm, diameter 1.68 fm).
Distribute the universe's holographic memory across space and one bit corresponds to exactly one nucleon's worth of volume. This does not mean "one proton is one bit", of course — the numbers simply happen to agree. Still, the coincidence stops you in your tracks.
04What scale is this?
the bare value without the coefficient
$$\left(R_H\,\ell_P^2\right)^{1/3}=3.24\ \mathrm{fm}\qquad(\times\,0.604\ \text{gives }1.96\ \mathrm{fm})$$So it is the "cube-root intermediate scale" between the horizon radius and the Planck length. Blend the largest and smallest lengths in the universe with these weights and out comes the size of a nucleon — an unexplained numerical coincidence, of the same kind as Extra 5's "\(\rho_\Lambda^{1/4}\) and the neutrino mass differ by only a factor 22".
Figure: how the three numbers grow as the universe grows. Cells go as slope 3, four-volumes as slope 4, bits as slope 2. Only the bits are slow, so the address shortage widens with size — they sufficed only while the universe was under 4.5 Planck lengths.
05The address table itself will not fit in memory
addressing every cell
$$5.27\times10^{182}\times607=3.20\times10^{185}\ \text{bits}$$against the memory
$$\frac{3.20\times10^{185}}{2.96\times10^{122}}=1.1\times10^{63}\ \text{times}$$Conclusion of §05
Merely addressing every cell would take \(10^{63}\) times the memory.
── The address table will not fit in memory. The address space exceeds anything the universe can handle.
This is a different shortage from Episode 6's occupancy of \(10^{-18}\). That was "the capacity is there and unused". This is "there are not enough addresses in the first place" — a problem prior to use.
06The time direction is a further 61 orders short
Space alone gives \(10^{-61}\); include time and it is \(10^{-122}\) — exactly twice the orders (as it must be, \(N\propto R^2\) against \(R^4\)). The meaning is clear: "recording the entire history of the universe" is impossible in principle.
07The reveal — this is not compression
Holography is sometimes described as "the information of a volume compressed onto an area". Read in the language of addresses, that phrasing misleads.
| If it were compression | Actual holography | |
|---|---|---|
| Original information | a volume's worth | only ever an area's worth |
| Operation | squeeze out redundancy | nothing is squeezed |
| Recovery | you can restore it | there is no "original" to restore |
| The right phrasing | ── | volume cells were never given addresses |
Episode 13 drew the line "a conformal transformation touches only size". Today's line is more basic — the universe as a storage device has its addresses set by area, not volume. So of \(10^{182}\) cells only \(10^{122}\) can be designated, and the gap widens as the universe grows.
① Counting "spatial cells" as \((R_H/\ell_P)^3\) is not a claim that spacetime is a discrete lattice. It is an indicative count in units of the Planck volume — the same metaphorical use as "cells / ticks" in Extra 3 of the previous series.
② The holographic bound is an inequality on what fits, not a guarantee of usable capacity (same caveat as Episode 6 ②). So "not enough addresses" is a statement about the bound, not a report of a failed attempt to record something.
③ There is no known explanation for \(\ell_{\rm bit}=1.96\) fm. It is not an identity (give \(R_H\) and \(\ell_P\) independently and it takes any value) and no mechanism is known. It belongs in the "coincidence" column. The observation that \((R_H\ell_P^2)^{1/3}\) lands on the nucleon scale is of a kind found in the literature and is not a discovery of this document.
④ The address-width argument of §05 assumes a naive encoding. In practice cells need no individual addresses (the coordinates are the address), and "building an address table" is not physically required — read it as a way of feeling the size of the address space.
⑤ \(R_H=ct_0\) is the \(c\cdot t=\text{const}\) convention. In \(\Lambda\)CDM, \(R_H=c/H_0\) differs from the particle horizon and the numbers shift by factors of a few — read this as an order-of-magnitude argument.
Exercises (solvable with this episode's formulas alone)
- Express the ratio of bits to spatial cells in terms of \(R_H/\ell_P\).
Show the answer
\(N/(R_H/\ell_P)^3=[\pi(R_H/\ell_P)^2/\ln2]/(R_H/\ell_P)^3=(\pi/\ln2)/(R_H/\ell_P)\). Proportional to \(1/R_H\), so the shortage worsens as the universe grows. - When were there enough addresses?
Show the answer
Solve \(N=(R_H/\ell_P)^3\): \(R_H/\ell_P=\pi/\ln2=4.53\). Only while the universe was smaller than 4.5 Planck lengths; never since. - Find the side of the volume one bit is responsible for.
Show the answer
\((R_H/\ell_P)^3/N=(\ln2/\pi)(R_H/\ell_P)=1.78\times10^{60}\) Planck volumes \(=7.52\times10^{-45}\ \mathrm{m^3}\). The cube root is 1.96 fm — the size of a proton. - What kind of scale is that?
Show the answer
\(\ell_{\rm bit}\propto(R_H\ell_P^2)^{1/3}\), the cube-root intermediate scale between horizon radius and Planck length. Bare value 3.24 fm, times the coefficient \((\ln2/\pi)^{1/3}=0.604\) gives 1.96 fm. An unexplained numerical coincidence — neither an identity nor physics. - (Harder) Why is calling holography "compression" misleading?
Show the answer
Compression would mean squeezing a volume's worth of information by removing redundancy, but in fact only an area's worth ever exists. Nothing is squeezed and there is no "original" to restore. The right phrasing is "volume cells were never given addresses" — a statement about the structure of the address space, not the amount of information.
Summary — addresses grow only with area
Three numbers: spatial cells \((R_H/\ell_P)^3=5.27\times10^{182}\), four-volume cells \(4.25\times10^{243}\), bits writable \(2.96\times10^{122}\). Exponents \(3\), \(4\), \(2\) — only the bits go as a square, which is the entire content of holography.
The ratio came out a clean identity: \(N/(R_H/\ell_P)^3=(\pi/\ln2)/(R_H/\ell_P)=5.61\times10^{-61}\). "Only \(10^{-61}\) of the spatial cells can be addressed", and since it goes as \(1/R_H\), the gap widens as the universe grows. Addresses sufficed only while the universe was under 4.5 Planck lengths.
Inverted, it gives the volume per bit — \(1.78\times10^{60}\) Planck volumes, side 1.96 femtometres. The size of a proton. It is \((R_H\ell_P^2)^{1/3}\), the cube-root intermediate between the largest and smallest lengths in the universe — neither identity nor physics, and for now a coincidence and nothing more.
Beyond that, the address table itself will not fit in memory (addressing every cell takes \(10^{63}\) times it). Counting time as well widens the shortage to \(10^{-122}\), so "recording the entire history of the universe" is impossible in principle. And finally a matter of words — holography is not compression. Nothing is squeezed and there is no original. Properly: "volume cells were never given addresses". Where Episode 6's \(10^{-18}\) was "there and unused", this was a shortage prior to use.
This document is Episode 18 of "c·t = const, That Clicks", written for physics-minded high-school and university readers. The holographic bound \(N=A/(4\ell_P^2\ln2)\) is standard. The values \((R_H/\ell_P)^3=5.27\times10^{182}\), \((ct_0/\ell_P)^4=4.25\times10^{243}\), \(N/(R_H/\ell_P)^3=(\pi/\ln2)/(R_H/\ell_P)=5.61\times10^{-61}\), \(1.78\times10^{60}\) Planck volumes per bit (side \(1.96\) fm), the 607-bit address width and the factor \(10^{63}\) for an address table, \(N/(ct_0/\ell_P)^4=7.0\times10^{-122}\), and "addresses sufficed only for \(R_H/\ell_P<\pi/\ln2=4.53\)" are all computed here (kenshou/calc22.py). Counting "spatial cells" in Planck volumes is indicative and not a claim that spacetime is a discrete lattice (the same metaphorical use as "cells / ticks" in Extra 3 of the previous series). The holographic bound is an inequality on what fits, not a guarantee of usable capacity, so "not enough addresses" is a statement about the bound. There is no known explanation for \(\ell_{\rm bit}\simeq1.96\) fm landing on the nucleon scale; it is neither an identity nor a mechanism — the observation that \((R_H\ell_P^2)^{1/3}\) lands there is of a kind found in the literature and is not a discovery of this document. The address-width argument in §05 assumes a naive encoding; in practice coordinates are the address and no table is physically required. \(R_H=ct_0\) is the \(c\cdot t=\)const convention; in \(\Lambda\)CDM \(R_H=c/H_0\) differs from the particle horizon and the numbers shift by factors of a few. Linear expansion is a minority model under examination and conflicts with nucleosynthesis when extrapolated into the early universe (Lewis, Barnes & Kaushik 2016). 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).