c·t = CONST, THAT CLICKS EPISODE 40 ／ Part V — where the tool breaks
Three "10^122" numbers turned out to be one
Not even
halfway along
What the universe has processed and what the universe can hold are the same number.
And what it actually uses is \(10^{-18}\) of that.
Episode 39 counted a solar-mass black hole as holding \(1.5\times10^{77}\) bits of entropy. Information about what, exactly? This time we count again for the whole universe — and find that three separate "\(10^{122}\)" headline numbers are in fact one and the same. The compression of Episode 26 works again.
01Gravitational entropy sits entirely in the weight-0 column
| Quantity | Weight | What it is |
|---|---|---|
| \(S/k_B=A/4\ell_P^2\) | \(+2-(+2)=\mathbf{0}\) | an area divided by an area |
| Bekenstein bound \(2\pi ER/\hbar c\) | \(-1+1-0=\mathbf{0}\) | energy × length |
| Weyl\(^2\)/Ricci\(^2\) (Penrose) | \(-4-(-4)=\mathbf{0}\) | curvature\(^2\) divided by curvature\(^2\) |
| \(K\cdot r_s^4\) (Kretschmann × radius\(^4\)) | \(-4+4=\mathbf{0}\) | used in §07 |
Conclusion of §01
The whole subject of gravitational entropy lives in the dimensionless column.
── As §03 of Episode 36 found: it reaches the arena where it can be judged.
02The Hubble sphere sits exactly at its own Schwarzschild radius
\(r_s/R_H=1.000000\)
Not a coincidence but an identity. On Episode 19's scale, 0 bits — nothing to be surprised by.
03The core — three "\(10^{122}\)" numbers are one number
| Quantity | Value |
|---|---|
| Holographic bound \(A/4\ell_P^2\) | \(2.265\times10^{122}\) nat \(=3.268\times10^{122}\) bit |
| \(S/k_B\) if all the mass were one black hole | \(2.265\times10^{122}\) nat |
| Episode 24's \(N=C\cdot t_0\) | \(3.107\times10^{122}\) bit |
and the gap is exactly \(t_0/(R_H/c)=0.9505\) — the difference between the age of the universe and the Hubble time
The main point of this episode
The three headline numbers were one.
"How much information the universe has processed" (Episode 24) and "how much information the universe can hold" (the holographic bound) are the same number,
and the 5 per cent gap is the ratio of the age to the Hubble time, nothing else.
── On Episode 19's scale, 0 bits. Episode 26's compression worked again.
04So what is the actual entropy?
| Component | \(S/k_B\) | Fraction |
|---|---|---|
| Supermassive black holes | \(3.10\times10^{104}\) | \(1.00\) |
| Stellar-mass black holes | \(2.20\times10^{96}\) | \(7.1\times10^{-9}\) |
| CMB photons | \(2.03\times10^{89}\) | \(6.6\times10^{-16}\) |
| Relic neutrinos | \(1.93\times10^{89}\) | \(6.2\times10^{-16}\) |
| Relic gravitational waves | \(2.3\times10^{87}\) | \(7.4\times10^{-18}\) |
| Stars, interstellar medium etc. | \(2.6\times10^{81}\) | \(8.4\times10^{-24}\) |
These are Egan & Lineweaver's (2010) estimates. 99.999… per cent of it is already gravitational entropy — horizon entropy. Everything that is not a black hole sums to \(3.8\times10^{89}\), which is 49.5 bits (15 orders of magnitude) below the black holes.
05How much of the capacity is in use?
| The ladder of doublings | \(S/k_B\) | \(\log_2 S\) |
|---|---|---|
| Around recombination (CMB + neutrinos) | \(3.96\times10^{89}\) | \(297.6\) |
| Today (total) | \(3.10\times10^{104}\) | \(347.1\) |
| The bound (holographic) | \(2.27\times10^{122}\) | \(406.5\) |
Conclusion of §05
49.5 doublings from the beginning until now; 59.3 doublings from now to the bound.
── The universe is not even halfway along.
Figure: the ladder of entropy doublings. The horizontal axis is \(\log_2 S\) — how many times the entropy has doubled. Move the slider to see where today's universe sits — the headroom is longer than the road already travelled.
06What the headroom is — gravity's degrees of freedom are not thermalised
07Penrose's candidate, and its weakness
| Black hole | \(r_s\) [m] | \(K\) [m\(^{-4}\)] | \(K\cdot r_s^4\) |
|---|---|---|---|
| Solar mass | \(2.954\times10^{3}\) | \(1.576\times10^{-13}\) | \(12.0\) |
| M87\(^*\) (\(6.5\times10^9\,M_\odot\)) | \(1.920\times10^{13}\) | \(8.828\times10^{-53}\) | \(12.0\) |
Conclusion of §07
The Weyl side is finite and dimensionless — at the horizon \(K\cdot r_s^4=12\), the same 12 whatever the mass.
What breaks is the dividing by Ricci.
── The ratio form works in cosmology but not in vacuum. Alternative definitions have been proposed (Clifton–Ellis–Tavakol 2013 among them).
(1) §04's figures are order-of-magnitude estimates. Egan & Lineweaver themselves assign a large uncertainty to the supermassive black hole entropy (it depends on extrapolating the mass function) — \(3.1\times10^{104}\) is a central value that can move by an order of magnitude either way, and §05's "59.3 doublings" moves with it.
(2) §03's "the same number" refers to counting on the Hubble sphere. Count on the particle horizon (about \(4.4\times10^{26}\) m) and the value is roughly ten times larger; the event horizon gives another answer — "which horizon?" has to be named or the number is not yet a sentence (the same structure as Episode 3 and Episode 37 §05). The identity \(r_s=R_H\) in §02 holds only for the Hubble sphere at critical density.
(3) §05's "bound" is the holographic bound, but which surface it should be applied to in an expanding universe is not obvious (Bousso's covariant entropy bound is one formulation) — "the universe uses \(10^{-18}\) of its capacity" is the statement for the naive form of the bound.
(4) §07's \(C^2/R^2\) is the most naive way of writing Penrose's idea. Penrose did not propose this ratio as a final definition; it is the widely quoted gauge for the idea that Weyl curvature measures gravitational entropy. Its divergence in vacuum is a weakness of the naive form, and later proposals try to repair exactly that.
(5) There is still no agreed definition of "the entropy of the gravitational field". The black-hole entropy \(A/4\ell_P^2\) is established, but the general case without a horizon is unresolved — §06 states the problem, not an answer.
Exercises
- What is the ratio of the Hubble radius to the Schwarzschild radius of the mass it contains?
Show the answer
Exactly 1. At critical density, \(r_s=\frac{2G}{c^2}\cdot\frac43\pi R^3\cdot\frac{3H^2}{8\pi G}=\frac{R^3H^2}{c^2}=R\). Not a coincidence but an identity, so 0 bits on Episode 19's scale — though per caveat (2) it holds only for the Hubble sphere at critical density. - Episode 24's \(N=C\cdot t_0\) is 0.95 of the holographic bound. What is the 5 per cent?
Show the answer
The ratio of the age of the universe to the Hubble time, \(t_0/(R_H/c)=0.9505\), exactly. So the two are the same number, and even the gap is accounted for — Episode 26's compression working again. - What fraction of the universe's entropy is in black holes?
Show the answer
99.999… per cent. Everything else sums to \(3.8\times10^{89}\), which is 49.5 bits (15 orders) below the total \(3.1\times10^{104}\). The entropy of the universe is already almost entirely gravitational. - Why is only \(10^{-18}\) of the capacity used?
Show the answer
Because gravity only attracts. A uniform gas is at maximum entropy, but a uniform gravitational field is at minimum entropy — the sign is reversed. So gravity's degrees of freedom cannot sit at thermal equilibrium, and headroom remains. - (Harder) What is the weakness of \(C^2/R^2\)? Put it in numbers.
Show the answer
It diverges in vacuum. Schwarzschild and Kerr have Ricci \(=0\), so the denominator vanishes. The Weyl side is healthy: at the horizon \(K\cdot r_s^4\) is 12.0 for both a solar-mass hole and M87\(^*\) — finite and dimensionless. Only the dividing by Ricci is broken.
Summary: the three headline numbers were one
The subject of gravitational entropy sits entirely in the weight-0 column — \(A/4\ell_P^2\), the Bekenstein bound, \(C^2/R^2\), \(K\,r_s^4\), all dimensionless.
Counting it out, the Hubble sphere sits exactly at its own Schwarzschild radius (\(r_s/R_H=1.000000\)) — an identity at critical density, worth 0 bits on Episode 19's scale. Because of it, three \(10^{122}\) numbers coincide: the holographic bound, the entropy of the total mass as one black hole, and Episode 24's \(N=C\cdot t_0\). "How much the universe has processed" and "how much the universe can hold" are the same number, and the 5 per cent gap is exactly the ratio of the age to the Hubble time. Episode 26's compression worked again.
And the actual entropy? \(3.1\times10^{104}\), of which 99.999… per cent is supermassive black holes — the entropy of the universe is already almost entirely gravitational. Against the bound that is \(1.4\times10^{-18}\), i.e. 59.3 doublings of headroom. Since the road from the beginning to now was 49.5 doublings, the universe is not even halfway along.
The headroom survives because gravity only attracts — a uniform gas is at maximum entropy, a uniform gravitational field at minimum. So "the entropy of the gravitational field itself" needs its own measure, and the candidate is the Weyl curvature. Penrose's naive form \(C^2/R^2\) has the property one wants — exactly zero for FLRW — but diverges in vacuum, where the denominator goes to zero. The Weyl side is healthy: at the horizon \(K\,r_s^4\) is 12.0 regardless of mass. Only the dividing by Ricci is broken.
This document is Episode 40 of "c·t = const, That Clicks" (the fourth of Part V), written for physics-minded high-school and university readers. Bekenstein–Hawking entropy, the holographic bound, and Penrose's idea of gauging gravitational entropy by Weyl curvature are all standard, and nothing here is a new claim — the numbers are computed in kenshou/calc44.py. §04's figures are from Egan & Lineweaver (2010, ApJ 710, 1825) and carry a large uncertainty on the supermassive black hole entropy (it depends on extrapolating the mass function); \(3.1\times10^{104}\) is a central value that can move an order of magnitude either way, and §05's "59.3 doublings" moves with it. §03's "the same number" refers to counting on the Hubble sphere; the particle horizon (about \(4.4\times10^{26}\) m) gives roughly ten times more — "which horizon?" must be named or the number is not yet a sentence. The identity \(r_s=R_H\) in §02 holds only for the Hubble sphere at critical density. §05's bound is the holographic one, but which surface it applies to in an expanding universe is not obvious (Bousso's covariant entropy bound is one formulation), so this is the statement for the naive form. §07's \(C^2/R^2\) is the most naive way of writing Penrose's idea and not a final definition he proposed — the vacuum divergence is a weakness of that naive form, which later proposals (Clifton–Ellis–Tavakol 2013 among them) try to repair. There is still no agreed definition of "the entropy of the gravitational field" away from horizons — §06 states the problem, not an answer. ── To make a PDF, use your browser's Print dialogue (sliders freeze and answers are hidden in the print version).