Ultra-locality generates \(S\propto A\) automatically ── and all that remains is the wall we measured in Episode 1
So far we have been chasing "how to show that it has \(e^{A/4G}\) states." Episode 10 found the template for the proof, Episode 11 re-placed the question.
This episode steps back and decomposes what actually has to be shown. The claim splits into two ── that it is proportional to the area, and that the constant of proportionality is \(1/4G\).
And the former comes out almost automatically from the properties of a Carrollian theory. Only the latter is hard. And that latter turned out to be the same wall this series has run into again and again since Episode 1.
We use what Episodes 5 and 8 confirmed.
In Carroll geometry the light cone is collapsed, so no signal propagates. As the figure in Episode 8 showed, the field equation becomes \(\partial_t^2\varphi=0\), with no spatial derivative on the right ── each point evolves in time without consulting its neighbours at all.
In a Carrollian theory, two events are causally connected only when they are at the same point. Each point of space is an independent system.
Statistical mechanics 101. Line up independent systems and the partition function is a product.
$$Z=\prod_{x}Z_x\qquad\Longrightarrow\qquad \log Z=\sum_x \log Z_x$$And in an ultra-local theory this \(x\) is each point of space. So \(\log Z\) is proportional to the number of points.
Now recall where a Carrollian theory lives. As Episode 7 confirmed ── on a null hypersurface. For a horizon, its spatial section is a two-dimensional surface.
ultra-local \(\Longrightarrow\) the entropy is proportional to the number of points
the points lie on a surface \(\Longrightarrow\) the number of points is proportional to the area
Hence \(S\propto A\).
That is what "comes out for free" means. No counting of black-hole microstates, no Cardy formula. The area law follows from nothing but an ultra-local theory living on a two-dimensional surface.
The area law is out, so the remainder is the coefficient. In formulae: cut the surface into cells of side \(a\), and let the number of states per cell be \(q\) ──
$$S=N\log q,\qquad N=\frac{A}{a^2}\qquad\Longrightarrow\qquad S=\frac{\log q}{a^2}\,A$$For this to match Bekenstein–Hawking \(S=A/4\ell_P^2\) ──
$$\frac{\log q}{a^2}=\frac{1}{4\ell_P^2}$$The important point here is that \(a\) and \(q\) are not determined individually. Only the combination is. Make the cells finer and the number of states per cell drops; coarser and it rises ── and the entropy does not change. Move it and check.
Move the cells and the cell count \(N\) and the states per cell \(q\) both change, yet the product \(N\log q\) does not. The area law is robust, and what is fixed is a single number: the information per unit area.
Convert \(S/k_B=A/4\ell_P^2\) into bits (divide by \(\log_2\))
$$\text{number of bits}=\frac{A}{4\ell_P^2\ln 2}=\frac{A}{2.773\,\ell_P^2}$$that is
\(4\ln2=2.773\) Planck areas per bit. Per unit area, \(1.381\times10^{69}\) bit/m².
| null surface | area [m²] | bits |
|---|---|---|
| solar-mass black hole | \(1.10\times10^{8}\) | \(1.51\times10^{77}\) |
| Sgr A* | \(2.03\times10^{21}\) | \(2.80\times10^{90}\) |
| de Sitter horizon | \(2.35\times10^{53}\) | \(3.24\times10^{122}\) |
Restating Episode 9's state count of the universe \(e^{2.25\times10^{122}}\) in bits gives this ── the universe is a \(3.24\times10^{122}\)-bit machine.
Let us put this episode into one line.
\(S\propto A\) came out of the structure. All that remained was the \(1/4\).
Section 08 of Episode 1 said this.
For black-hole entropy: dimensions get you as far as \(S\propto A\), and only the \(1/4\) refuses to come. Filling that in requires the thing being counted itself ── states, a mechanism, a symmetry.
We have arrived at exactly the same place. And this time \(S\propto A\) came out of a structure, not even from dimensional analysis. And still the \(1/4\) does not come.
The \(O(1)\) wall this series has been measuring since Episode 1 stands at the same height even when you change route.
This has to be written honestly. §02's argument has a hidden hole.
A continuum has infinitely many points.
We said "proportional to the number of points," but a continuous surface has uncountably many. So the moment we wrote \(N=A/a^2\), we put in a cutoff \(a\) by hand. And \(\log q/a^2\) can diverge as \(a\to0\).
The divergence of entanglement entropy comes from tracing out the degrees of freedom outside the surface (in the bulk). Erase the outside and infinity accumulates at the seam.
The Carrollian case is different. The theory itself is defined on the surface (Episodes 7 and 8). There is no "outside" to trace out.
Then the coefficient \(\log q/a^2\) should be fixed not by an external cutoff but by the theory's own density of degrees of freedom. And that is ──
density of degrees of freedom per unit area = density of states = Episode 11's "central charge" question.
The problem has not disappeared. It has merely moved ── the same shape as Episode 3's "the curvature did not vanish; it moved into the conformal factor."
But it has moved somewhere better. From "cutoff-dependent and undetermined in principle" to "should be fixed as data of the theory, but we cannot compute it yet." From a question with no answer to a question with one.
The area law comes out of the structure automatically. If ultra-local, the partition function is a product over points and \(\log Z\) is proportional to the number of points. And a Carrollian theory lives on a null surface, so the number of points is proportional to the area ── \(S\propto A\). No counting of microstates, no Cardy.
Whereas ordinary field theory gives a volume law. The area law was long a puzzle because it was a departure from the volume law. In the Carrollian case the place the theory lives is a surface to begin with, so it is the default rather than a departure.
The coefficient condenses into a single number. The cell size \(a\) and the states per cell \(q\) are not determined individually; only \(\log q/a^2\) is. That it equals \(1/4\ell_P^2\) means ── \(4\ln2=2.773\) Planck areas per bit, \(1.381\times10^{69}\) bit/m². For the de Sitter horizon, \(3.24\times10^{122}\) bits.
And back to Episode 1 for the third time. "Dimensions get you as far as \(S\propto A\), and only the \(1/4\) refuses" ── this time \(S\propto A\) came out of a structure rather than dimensional analysis, and the wall still stood at the same height.
But only the form came for free. A continuum has infinitely many points, so \(N=A/a^2\) puts in a cutoff by hand ── the same divergence as the entanglement entropy \(S\sim A/\epsilon^2\) of ordinary QFT. Though in the Carrollian case there is no "outside" to trace out, so the coefficient should be fixed by the theory's own density of degrees of freedom. That is exactly Episode 11's central-charge question, so we have gone full circle.
This document is Episode 12 of the "Lattice We Build" series, a reading piece for high-school and university students who love physics. Where the sister series "That Clicks" explains known physics, this series shows the work itself.
Established material: the ultra-locality of Carrollian theories; that the partition function of independent systems is a product; the Bekenstein–Hawking entropy \(S=A/4\ell_P^2\) and the numbers computed from it (\(4\ln2=2.773\,\ell_P^2\) per bit, \(1.381\times10^{69}\) bit/m², the bit counts of each null surface); and the area law and UV divergence of entanglement entropy in ordinary field theory.
On the other hand, §02's argument that "the black-hole area law is derived from the ultra-locality of a Carrollian theory" is this series' author's structural framing, not an established derivation. Indeed, as §05 states, a continuum has infinitely many points and cutting into cells amounts to introducing a cutoff. The coefficient \(1/4\) is not derived by this argument, and whether the density of degrees of freedom that would give it is fixed as data of the theory is also unsolved. That the horizon's Carrollian theory itself is unconstructed is as stated in Episodes 8 and 11.
Main series: Episode 1｜Episode 2｜Episode 3｜Episode 4｜Episode 5｜Episode 6｜Episode 7｜Episode 8｜Episode 9｜Episode 10｜Episode 11 ｜ bonus: ①／②／③ ── to print, use your browser's "Print" → "Save as PDF."
Print / PDF: ⌘+P (Ctrl+P on Windows). In the figure you can confirm that changing the cells does not move the entropy.