Pushing route B answered Episode 12's question ── except the answer is "that question is not needed"
Episode 12 wrote that the area law itself comes out of the Carrollian structure while only the coefficient does not. It concluded "the coefficient should be fixed by the theory's own density of degrees of freedom" and thereby returned to Episode 11's central-charge question ── going full circle.
This time we cut that circle from another side. Route B obtains the free energy from a local effective action, so it is precisely a tool for computing that "density of degrees of freedom." If usable, Episode 12's homework is done.
The conclusion first. Route B does not work in four dimensions. But pinning down why shows that the very way Episode 12 set up the question was wrong.
Two-dimensional Cardy came from modular invariance (Episode 10, §03). Exchange the two cycles of the torus and high and low temperature swap ── that symmetry constrains the state count.
Route B does not use that symmetry. Instead ──
Put a system at temperature \(1/\beta\) on a space with a thermal circle of circumference \(\beta\). As \(\beta\to0\) that circle collapses, so the theory reduces to one dimension lower. Read off the coefficients of the reduced local effective action and you get \(\log Z\) at high temperature.
Since no modular invariance is needed, it does not care about the dimension. This is the line Di Pietro–Komargodski followed for supersymmetric theories in \(d=4,6\).
Again, check the tool first. Run route B in two dimensions and see whether it gives the same answer as Cardy.
The route-B side is just thermodynamics. The high-temperature free energy of a 2d CFT is \(F=-\pi cL/(6\beta^2)\), so ──
$$S=-\frac{\partial F}{\partial T}=\frac{\pi cL}{3\beta}$$The route-A side is Cardy. Get \(L_0\) from the energy and put it in.
# B: thermal S = pi c L / (3 beta) # A: Cardy S = 2pi sqrt(c L0/6) + 2pi sqrt(c L0bar/6), L0 = L0bar = E L/(4pi) c L beta | S_thermal S_Cardy ratio -------------------------------------------------------------------- 1 10 0.3 | 34.90658504 34.90658504 1.000000000000000 24 100 0.5 | 5026.54824574 5026.54824574 1.000000000000000 0.5 7 0.05 | 73.30382858 73.30382858 1.000000000000000 3.7 1000 2.1 | 1845.06235211 1845.06235211 1.000000000000000 max deviation 2.2e-16 → positive control PASSED
An exact match. But it matches too well, and that is the point.
The leading coefficient of the thermal free energy is the central charge \(c\), directly.
That is, in two dimensions "the number of degrees of freedom" and "the central charge" are the same number. Which is why Cardy works ── the quantity to be counted happened to be a universal one.
Here is the wall. Do the same in four dimensions and the leading term is
$$\log Z\;\sim\;\frac{V}{\beta^{3}}\times(\text{coefficient of the free-energy density})$$And this coefficient changes with the coupling. Let us check it in the most famous example ── \(\mathcal{N}=4\) supersymmetric Yang–Mills theory.
The free-field free-energy density is \(f=-\dfrac{\pi^2T^4}{90}\left(n_b+\tfrac78 n_f\right)\). For \(\mathcal{N}=4\), \(n_b=n_f=8N^2\), so
$$f_{\text{free}}=-\frac{\pi^2T^4}{90}\left(8+\tfrac78\cdot8\right)N^2=-\frac{\pi^2N^2T^4}{6}$$The same theory, and changing the coupling multiplies the coefficient by \(3/4\). "The number of degrees of freedom" is a function of the coupling.
This is fatal. What we want is \(A/4G\) ── a universal number independent of the matter content. A universal quantity does not come out of a non-universal one.
That does not make route B useless. Use a supersymmetric index and this non-universal leading term disappears.
Weight by \((-1)^F\) and long representations cancel, leaving only BPS states. Then the \(\beta^{-3}\) volume term drops entirely ── in the papers' phrasing, "a supersymmetric theory generates no cosmological constant."
And the next term, which shows its face after the drop, was universal.
\(a,c\) are the trace-anomaly coefficients. \(\mathrm{Tr}\,R=16(a-c)\).
Anomaly coefficients do not move with the coupling ('t Hooft anomaly matching). So this is a genuine universal quantity. Let us check it on a free chiral multiplet.
# a = (3/32)(3 TrR^3 - TrR), c = (1/32)(9 TrR^3 - 5 TrR) identity a - c = TrR/16 TrR^3=1 TrR=0 -> a=9/32 c=9/32 a-c=0 TrR/16=0 OK TrR^3=0 TrR=1 -> a=-3/32 c=-5/32 a-c=1/16 TrR/16=1/16 OK TrR^3=-1/27 TrR=-1/3 -> a=1/48 c=1/24 a-c=-1/48 TrR/16=-1/48 OK free chiral multiplet (R=2/3, the fermion has R=-1/3) a = 1/48 known value 1/48 : OK c = 1/24 known value 1/24 : OK DP-K: log Z = -16 pi^2 (a-c)/(3 beta) = pi^2/(9 beta) N=4 SYM SU(N) (1 vector + 3 chirals, all adjoint) N=2 TrR=0 a=3/4 c=3/4 a==c known (N^2-1)/4 = 3/4 OK N=3 TrR=0 a=2 c=2 a==c known (N^2-1)/4 = 2 OK N=10 TrR=0 a=99/4 c=99/4 a==c known (N^2-1)/4 = 99/4 OK => in N=4 SYM, a=c. the DP-K universal term is exactly zero.
The formulae are right, and they reproduce the known \(a=1/48,\;c=1/24\) for a free chiral. But ──
Stopping plainly here ends at "route B failed too." Before stopping, there is one thing that nags.
Let us restate the fact seen in §03.
| quantity | does it depend on the matter content? |
|---|---|
| the free-energy coefficient of the field theory (= the number of degrees of freedom) | yes (a factor \(3/4\) in \(\mathcal{N}=4\)) |
| the coefficient of the entanglement entropy \(S\sim A/\epsilon^2\) | yes (grows with the number of field species) |
| the black-hole entropy \(A/4G\) | no |
Read normally, this is a contradiction. Add one more field species and the entanglement across the horizon certainly increases. And yet \(A/4G\) does not move. Where did the increase go?
This is the long-standing difficulty called the species problem. It is the same divergence that Episode 12 honestly wrote about as "\(N=A/a^2\) puts in a cutoff by hand."
And the answer had already appeared, in 1994.
Susskind–Uglum's proposal is this ── the divergence of the entanglement entropy and the renormalization of Newton's constant are the same divergence.
Plausible as a statement, but the question is whether the coefficients match. Let us compute both sides independently and compare.
Solodukhin's review, eq. (28), gives the value for a single free scalar crossing a surface of area \(A\) ──
$$S_{EE}=\frac{A}{6(d-2)(4\pi)^{(d-2)/2}\epsilon^{\,d-2}} \;\xrightarrow{\;d=4\;}\;\frac{A}{6\cdot2\cdot4\pi\,\epsilon^2}=\frac{A}{48\pi\epsilon^2}$$This one is obtained from the heat kernel, without looking at entropy at all. The divergent part of the one-loop effective action of a single scalar, in the Seeley–DeWitt expansion, is
$$\Gamma_{\text{div}}=-\frac12\int_{\epsilon^2}^{\infty}\frac{ds}{s}\,\frac{1}{(4\pi s)^2}\int d^4x\sqrt{g}\;\bigl(a_0+a_1 s+\cdots\bigr), \qquad a_0=1,\;\; a_1=\Bigl(\tfrac16-\xi\Bigr)R$$This \(\tfrac16\) is the textbook coefficient and has nothing to do with entropy. Picking out only the \(R\) term ──
$$\Gamma_{\text{div}}\supset-\frac{1}{32\pi^2}\Bigl(\tfrac16-\xi\Bigr)\frac{1}{\epsilon^2}\int d^4x\sqrt{g}\,R$$Compare with the Euclidean Einstein–Hilbert action \(-\frac{1}{16\pi G}\int\sqrt{g}\,R\) ──
$$\frac{1}{G_{\text{ren}}}=\frac{1}{G_0}+\frac{1}{2\pi\epsilon^2}\Bigl(\tfrac16-\xi\Bigr) \qquad\xrightarrow{\;\xi=0\;}\qquad \frac{1}{G_{\text{ren}}}=\frac{1}{G_0}+\frac{1}{12\pi\epsilon^2}$$# right: entanglement entropy (from Solodukhin eq. 28) S_EE = A / (48 pi eps^2) # left: renormalization of G (from the heat kernel a1 = (1/6 - xi)R. derived without looking at entropy) delta(1/G) |_{xi=0} = 1 / (12 pi eps^2) A/4 * delta(1/G) = A / (48 pi eps^2) ------------------------------------------------ Agreement. And exactly, down to the factor of "4". # does it stay linked for general xi? xi = 0 delta(1/G) = (1/12)/(pi eps^2) required S_EE = A/(48 pi eps^2) xi = 1/12 delta(1/G) = (1/24)/(pi eps^2) required S_EE = A/(96 pi eps^2) xi = 1/6 delta(1/G) = 0 required S_EE = 0 xi = 1/4 delta(1/G) = -(1/24)/(pi eps^2) required S_EE = negative at conformal coupling xi=1/6, both sides cross zero simultaneously.
They match. And down to the factor of \(A/4\). One is "entanglement across a surface," the other "the coefficient of the \(R\) term in the gravitational effective action" — entirely different calculations. And they give the same number.
The density of degrees of freedom \(\sigma\) never had to be determined.
Add matter and \(S_{EE}\) increases. And the same matter increases \(1/G\) by exactly as much. So
$$S=\underbrace{\frac{A}{4G_0}}_{\text{bare}}+\underbrace{\frac{N_s A}{48\pi\epsilon^2}}_{\text{matter entanglement}}=\frac{A}{4G_{\text{ren}}}$$The form does not change. What is determined is not \(\sigma\) alone but only the ratio of \(\sigma\) to \(G\). Episode 12's "the cell side \(a\) and the states per cell \(q\) are not determined individually; only \(\log q/a^2\) is" ── was exactly this, all along.
Increase the species with the slider and the breakdown moves, but the total and the rightmost bar are always at exactly the same height. The split of "where geometry ends and matter begins" carries no meaning ── because you can move it.
Honestly: this has not derived the \(1/4\). It has only explained "why adding matter does not move the \(1/4\)"; \(G\) itself is still a number supplied from outside.
There is an interesting limit, though. The figure's "Sakharov" button. Set the bare \(1/G_0\) to zero ── the idea that gravity is not a fundamental force but wells up entirely out of matter loops (Sakharov 1967, induced gravity).
Then \(A/4G\) becomes entirely matter entanglement. And the required number of species is fixed ──
Set \(1/G_0=0\) and take the cutoff at the Planck length \(\epsilon=\ell_P\). With \(\hbar=c=1\), \(\ell_P^2=G\), so
$$\frac{1}{G}=\frac{N_s}{12\pi\epsilon^2}=\frac{N_s}{12\pi G}\qquad\Longrightarrow\qquad N_s=12\pi\simeq37.7$$About 38 scalars' worth. The order of magnitude matches the number of Standard Model degrees of freedom (roughly 100).
| result | |
|---|---|
| route B itself | does not work in four dimensions. the leading term is non-universal, DP-K's universal term requires a supersymmetric index, and it degenerates for \(\mathcal{N}=4\) |
| Episode 12's homework | solved. but in the form "no need to compute it" rather than "compute the density of degrees of freedom" |
| the \(1/4\) itself | still not obtained. the bare \(1/G_0\) remains an external input |
Episode 1's wall was there a fourth time. But its character has shifted slightly ── previously it was "there is no tool to produce the coefficient"; this time we learned that "the coefficient can never come out of matter degrees of freedom."
The counting direction has closed. One door closed in Episode 13, and another closed here. What remains is only the question of where to get \(G\) itself.
Positive control: in two dimensions, routes A and B were the same thing. \(S=\pi cL/3\beta\) from the thermal free energy agrees with Cardy to a deviation of \(2.2\times10^{-16}\). In two dimensions "the number of degrees of freedom" is the central charge itself, which is why Cardy works.
In four dimensions that leading term is not universal. The free-energy coefficient of \(\mathcal{N}=4\) SYM is \(-\pi^2N^2T^4/6\) at weak coupling and \(-\pi^2N^2T^4/8\) at strong ── a factor of \(3/4\). Even in the same theory, "the number of degrees of freedom" moves with the coupling. The universal \(A/4G\) will not come from here.
DP-K's rescue has a blind spot. With a supersymmetric index the \(\beta^{-3}\) term vanishes and the universal term \(\log Z\sim-\frac{16\pi^2}{3\beta}(a-c)\) appears (verified by reproducing \(a=1/48,\,c=1/24\) for a free chiral, and \(a-c=\mathrm{Tr}R/16\)). But it degenerates for \(\mathcal{N}=4\), where \(a=c\) ── nothing can be said in precisely the theory with a black-hole dual.
And the question had been set up wrong. Add matter and \(S_{EE}\) increases. But the same matter increases \(1/G\) too. The entanglement side \(A/48\pi\epsilon^2\) (Solodukhin eq. 28) and the \(\delta(1/G)=1/12\pi\epsilon^2\) obtained independently from the heat kernel (from \(a_1=(\frac16-\xi)R\)) agree exactly, down to the factor of \(A/4\). At \(\xi=1/6\) both sides cross zero simultaneously.
Episode 12's homework was solved with the answer inverted. The density of degrees of freedom \(\sigma\) need not be determined ── what is determined is only the ratio of \(\sigma\) to \(G\). Episode 12's "\(a\) and \(q\) are not determined individually; only \(\log q/a^2\) is" was exactly this. Not a circle: it closed.
And still the \(1/4\) is not obtained. The bare \(1/G_0\) remains an external input. Take Sakharov's induced gravity with \(1/G_0=0\) and everything becomes matter entanglement, requiring \(N_s=12\pi\simeq37.7\) species at \(\epsilon=\ell_P\) ── the order matches the Standard Model, but nothing more can be said, since fermions and gauge fields with different coefficients have not been mixed in. Episode 1's wall, a fourth time. But its character has changed ── from "there is no tool to produce the coefficient" to "the coefficient can never come out of matter degrees of freedom."
This document is Episode 14 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 ── including the tools that did not work and what was learned from that.
Established material: the agreement of the high-temperature free energy of a 2d CFT with the Cardy formula; the free-field free-energy density \(f=-\frac{\pi^2T^4}{90}(n_b+\frac78n_f)\); that the strong-coupling free energy of \(\mathcal{N}=4\) SYM is \(3/4\) of the free-field value (Gubser–Klebanov–Peet); the relation of \(a,c\) to the anomaly coefficients in \(\mathcal{N}=1\) and that \(a=c=(N^2-1)/4\) for \(\mathcal{N}=4\); Di Pietro–Komargodski's \(d=4\) Cardy-type formula (arXiv:1407.6061); that the Cardy limit of the \(\mathcal{N}=4\) index requires complex chemical potentials to reproduce the AdS₅ black hole; the Seeley–DeWitt coefficient \(a_1=(\frac16-\xi)R\); the leading divergence of the entanglement entropy (Solodukhin, Living Rev. Relativity 14 (2011) 8, eq. 28); Susskind–Uglum's renormalization proposal; and Sakharov's induced gravity.
The computational parts of this article (the 2d A/B agreement, the anomaly-coefficient checks, the comparison of \(\delta(1/G)\) from the heat kernel with \(S_{EE}\), and \(N_s=12\pi\)) were carried out and checked by the author from the above. On the other hand, the reading that "therefore Episode 12's density-of-degrees-of-freedom question was set up wrongly" is this series' author's framing. The Susskind–Uglum proposal itself is widely accepted, but a proof that the divergences cancel exactly for general field content is not treated here (non-minimal couplings and higher-derivative terms require separate discussion). The \(N_s\simeq37.7\) of §08 is no more than an order-of-magnitude match and ignores the coefficient differences for fermions and gauge fields. That the horizon's Carrollian theory itself is unconstructed is as stated in Episodes 8, 11 and 12.
Main series: Episode 1｜Episode 2｜Episode 3｜Episode 4｜Episode 5｜Episode 6｜Episode 7｜Episode 8｜Episode 9｜Episode 10｜Episode 11｜Episode 12｜Episode 13 ｜ bonus: ①／②／③ ── to print, use your browser's "Print" → "Save as PDF."
Print / PDF: ⌘+P (Ctrl+P on Windows). The figure lets you confirm that the total does not move when you add matter.