[Paper Review] Bekenstein - Universal upper bound on the entropy-to-energy ratio for bounded systems
Original paper is here.
Another proof that use QFT is here.
Wikipikea is here.
Next paper is here.
1. Introduction
This paper make upper bound $$\begin{align}\frac{S}{E}< \frac{2\pi R}{\hbar c}\end{align}$$ for arbitrary sphere. Also shows black hold get maximum entropy.
We can make this question:
"Can we contentrate information infinitly?"
2. Why a bound on $S/E$?
Let the system have energy $E$ (or less than $E$). If the system is also bounded in space, then its phase space is bounded, and so must its entropy. See directly is difficult.
Kerr black hole
Before going on, we see how dependent is the bound (1) on the assumption that the self-gravity of the system is negligible? To clarify matters, consider a Kerr black hole, a system with maximal gravitational effects. Let its energy be $E$; then its surface area will be $$\begin{align}A=4\pi \{[E+(E^2-a^2-Q^2)^{1/2}]^2+a^2\},\end{align}$$ where $a$ is the specific angular momentum and $Q$ the charge. We define the effective radius $R$ by $4\pi R^2=A$. Then it is clear that $R<2E$, the equality corresponding to $a=Q=0$. Now since $S=A/4$, $S/ER=\pi R/E$; it then follows from the inequality that Kerr holes conform to the bound (1); the Schwarzschild hole actually attains the bound.
$$\begin{align}R=(A/4\pi)^{1/2}.\end{align}$$
$$\begin{align}S/E=4/3T,\end{align}$$
Low temperature limit
Gibbons has developed a microcanonical approach for determining the maximal $S/E$ for such systems. He labels the equantum states of the field by the occupation numbers of the various eigenfrequencies and takes the energy $E$ of each state to be the appropriate occupation numbers. He then assumes the entropy at a given $E$ to be given by the Boltzamnn formula $$\begin{align}S_{MC}(E)=\ln N(E),\end{align}$$ where $N(E)$ is the number of distince states with energy $E$. $N(E)$ can always be computed explicitly, at least for low $E$, if the eigenfrequency spectrum is known. But thus far no general method fort doing this has emerged and $N(E)$ must be computed separately and laboriously for each spectrum. In all cases for which this has been done (i.e., scalar field confined to a one-dimensional cavity, to the flat interior of a two-sphere, or to an Einstein universe $S^3$), $\ln N(E)/E$ has a maximum at some moderately low $E$. The maxima all lie below the bound (1) (with $R$ interpreted as the radius of the appropriate $n$-sphere) thus providing support for it.
These results also support our remarks about the relation between the bound on $S/E$ and the breakdown of the thermodynamic (continuum) description of a field at low temperatures (energies). They suggest that the maximal $S/E$ is a phenomenon of the low-excitation states of field. This may explain why the existence of the bound (1) has not been widely noticed. When cavities having two or three very different scales of length are considered in the Gibbons formalism, it becomes clear that the maximal $S/E$ is determined by the longest scale (that most clearly associated with the lowest eigenfrequencies). This feature is the primary motivation for the definition of effective radius (3): twice $E$ is never less than the longest dimension of the system.
Quantum Statistics
A quantum system is described its Hamiltonian $\bar{H}$; its state is described by a density operator $\hat{\rho}$ satisfying the normalization condition $$\begin{align}\operatorname{Tr}\hat{\rho}=1.\end{align}$$ The mean energy is $$\begin{align}\bar{E}=\operatorname{Tr}(\hat{\rho}\hat{H})\end{align}$$ and the entropy $$\begin{align}S=-\operatorname{Tr}(\hat{\rho}\ln\hat{\rho}).\end{align}$$
It is well known that the microcanonical entropy (5) is just the maximal value which $S$ attains over all $\hat{\rho}$ which satisfy (6) and which assign nonvanishing probabilities only to states with energy $E$. Thus $$\begin{align}\ln N(E)/E=\max_E (S/\bar{E}),\end{align}$$ with the maximization carried out under the same conditions as before. In particular, (9) holds for that $E$ for which $\ln N(E)/E$ is largest. If we now remove the constraint that $\hat{\rho}$ is confined to a given $E$, $\max (S/\bar{E})$ can only increase. Therefore $$\begin{align}\max [\ln N(E)/E]< \max (S/\bar{E}),\end{align}$$ where the maximization in the left-hand side is over $E$, while that in the right-hand side is over all $\hat{\rho}$ which fulfill (6). Thus the peak $S_{MC}/E$ value is bounded by the maximum that $S/\bar{E}$ attains for an arbitrary normalized $p$. If it can be shown that the last quantity is below the bound (1), then one will know that $S_{MC}/E$ fulfills it also.
3. Bound on $S/\bar{E}$ from quantum statistics
For simplicity we consider only systems composed exclusively of massless fields: scalar, electromagnetic, and neutrino (for technical reasons we do not consider gauge or metric fields). We ignore interactions, except as needed in enforcing boundary conditions at cavity walls. We assume the systems are stationary; this is clearly not a great restriction, for stationarity often implies equilibrium and maximal $S$, other things being equal. We consider fields in flat spacetime confined by cavities of arbitrary shape, or fields confined in a model (Einstein) universe of constant space curvature. We do not, however, consider fields in flat spacetimes with non-Euclidean topologies.
Let us ask which $\hat{\rho}$ makes $S/\bar{E}$ maximal? It is evidently given by the solution of the variational problem $$\begin{align}\delta [-\operatorname{Tr}(\hat{\rho}\ln\hat{\rho})/\operatorname{Tr}(\hat{\rho}\hat{H})-\lambda\operatorname{Tr}(\hat{\rho})]=0,\end{align}$$ where $\lambda$ is a Lagrange multiplier used to enforce the normalization condition (6). Varying $\hat{\rho}$, rearranging terms and normalizing, one fields $$\begin{align}\hat{\rho}=Z(\beta_0)^{-1}\exp (-\beta_0 \hat{H}),\\ Z(\beta_0)\equiv \operatorname{Tr} [\exp (-\beta_0 \hat{H})],\\ \beta_0=(S/\bar{E})_{max}.\end{align}$$ Thus we wamted distribution is the canonical one whose inverse temperature parameter is just the peak value of $S/\bar{E}$ for the system. To find this value one computes $S/\bar{E}$ with (12) and compares with (14) to get $$\begin{align}\ln Z(\beta_0)=0.\end{align}$$
Let $\omega_i$ be the eigenfrequencies for our system and let them be $g_i$-fold degenerate. Then at inverse temperature $\beta$ the mean energy in all species is given by $$\begin{align}\bar{E}=E_0+\sum_i g_i\omega_i(e^{\beta\omega_i}\mp 1)^{-1},\end{align}$$ where upper (lower) singns are for boson (fermion) modes. The $E_0$ is the sum of regularized vacuum energies, while the mode sum is just the usual thermal contribution. That the intuitive form (16) follows from detailed regularization of the finite-temperature quantum field theory. Substituting (16) and (12) into (7) and integrating with respect to $\beta$, one obtains the finite expression $$\begin{align}\ln Z=C-\beta E_0+\sum_i \mp g_i \ln (1\pm e^{-\beta\omega_i}),\end{align}$$ where $C$ is a constant of integration. Now using (12) to calculate $S$, one finds that at zero temeprature $(\beta\rightarrow \infty)$ $S\rightarrow C$ provided no $\omega_i$ vanishes (see Appendix C). However, for the fields we ahve in mind the vacuum state is non-degenerate; thus we expect $S\rightarrow 0$ at zero temperature (third law of tehrmodynamics). Hence, we must set $C=0$.
A. Case with $E_0>0$
Assume, as is probably the case, that the vacuum energy of every species of field confined to a finite cavity, or to a compect universe is positive; let $E_{0k}$ be the vacuum energy for the $k$th field. Now solve the equation $$\begin{align}E_{0k}=\beta_k^{-1}H_k(\beta_k)\equiv \mp \beta_k^{-1}\sum^{(k)}_i g_i \ln (1\mp e^{-\beta_k\omega_i})\end{align}$$ (which involves the sum over the $k$th field's modes only) for the parameter $\beta_k$. Repeat for every other species. One has then a series of distinct $\beta_k$. Now $E_0=\sum_k E_{0k}$. Therefore, substituting the expressions (18) for all $k$ into (17) evaluated at $\beta=\beta_0$ we get $$\begin{align}\beta_0^{-1}\sum_k H_k(\beta_0)=\sum_k \beta_k^{-1}H_k(\beta_k).\end{align}$$ Now, as direct differentiation shows, $\beta^{-1}H_k(\beta)$ is monotonic decreasing in $\beta$ both for fermions and bosons. It is then fairly clear from (19)that $\beta_0$, is bracketed by the smallest and largest $\beta_k$. This leads to great simplification: By treating field species separately one can find upper and loiter bounds for $\beta_0$ for any conceivable mixture of species.
1. Upper bound
Introduce $n_k(\omega)$, the number of modes (counting degeneracies) of species $k$ with eigenfrequencies up to $\omega$. Clearly at every $\omega$ which is an eigenvalue, $n_k(\omega)$ has a step discontinuity of strength $+g_I$. This (18) can be written as $$\begin{align}E_{0k}=\mp \beta_k\int^\infty_0 (dn_k/d\omega)\ln(1\mp e^{-\beta_k\omega})d\omega.\end{align}$$ Integration by parts gives $$\begin{align}E_{0k}=\int^\infty_0n_k(\omega)(e^{\beta_k\omega}\mp 1)^{-1}d\omega\end{align}$$ as a condition for $\beta_k$. In most cases we know little about the precise from of $n_k$. However, one can easily obtain from (21) an upper bound for $\beta_k$ which will suffice for our purpose - establishing that (1) holds generally.
One starts with the obvious inequality (valid for $p>0$) $$\begin{align}n_k(\omega)\omega^{-p}<\sum^\omega g_i\omega_i^{-p},\end{align}$$ where the sum includes all eigenfrequencies up to $\omega$. The inequality holds (for any $p>0$) since every $\omega_i<\omega$, while $n_k(\omega)$ is just the number of terms in the sum. The sense of the inequality is clearly preserved if which case it becomes the well-known $\zeta$ function for the Hamiltonian of the field $k$ (see also Appendix B), $$\begin{align}n_k(\omega)<\zeta_k(p)\omega^p.\end{align}$$ Since $\zeta_k(p)$ converges only for $p>3$, we restrict all our following remarks to that range. We now use (23) to set an upper bound on the integral in (21): $$\begin{align}E_{0k}<\beta_k^{-p-1}\zeta_k(p)\int^\infty_0 x^p(e^x\mp 1)^{-1}dx.\end{align}$$ Evaluating the integral in terms of Riemann's $\zeta$ function $\zeta_R(p)$ and isolating $\beta_k$, we get the bound $$\begin{align}\beta_k<[\Gamma(1+p)\zeta_R(1+p)\zeta_k(p)E_{0k}^{-1}(1-\delta_k2^{-p})]^{1/1(1+p)},\end{align}$$ where $\delta_k=1$ for fermions and $\delta_k=0$ for bosons.
For the scalar, electromagnetic, and neutrino fields in an Einstein universe one knowns all quantities in (25). For example, for the scalar field $E_{0k}=(240a)^{-1}$ and $\zeta_R(p)=a^p\zeta_R(p-2)$, where $a$ is the radius of the universe. By trial and error one finds that $p=6.4$ gives nearly the lowest bound on $\beta_k$. Inserting numbers one gets $$\begin{align}\beta_k(\mbox{scalar}; S^3)<5.70a.\end{align}$$ This may be compared with the exact value $\beta=4.13a$ which is obtained by finding (numberically) the zero of $\ln Z$. In liker manner and still using $p=6.4$, one finds $$\begin{align}\beta_k(\mbox{neutrino}; S^3)<4.25a,\end{align}$$ which again compares favorably with the exact result $\beta=2.61a$ inferred from the $\ln Z$. Finally, for the electromagnetic field one has (also with $p=6.4$) $$\begin{align}\beta_k(\mbox{electromagnetic}; S^3)<2.70a.\end{align}$$ In this case there exists no exact result for comparison. Evidently, if we identify $a$ with the effective radius of the universe $R$, then all our $\beta_k$'s are below $2\pi R$. Consequently, an Einstein universe filled with any mixture of photons, neutrinos and scalar particles obeys (1).
$$\begin{align}\beta_k \omega_i<[\Gamma(1+p)g_1\omega_1/E_{0k}]^{1/(1+p)}.\end{align}$$
$$\begin{align}\xi=\Gamma(1+p_c)\exp [-(1+p_c)\psi(1+p_c)]\end{align}$$
$$\begin{align}[\Gamma(1+p_c)\xi^{-1}]^{1/(1+p_c)}=\exp [\psi(1+p_c)],\end{align}$$
$$\begin{align}\psi(p)\equiv d\ln \Gamma(p)/dp.\end{align}$$
$$\begin{align}\beta_k\omega_1<13.30.\end{align}$$
$$\begin{align}\beta_k(\mbox{scalar or neutrino; cavity)}<4.233R\end{align}$$
$$\begin{align}\beta_k(\mbox{electromagnetic; cavity)}<6.388R.\end{align}$$
2. Lower bound
B. Case with $E_0=0$
IV. Assessment
Appendix B
Here we show by example that for large $p$ various $\zeta$ functions of interest are well approximated by their ground-state parts. For a spectrum $\omega_i$ with degeneracy $g_i$, $$\zeta(p)=\sum^\infty_{i=1}g_i\omega_i^{-p}.$$ The ground-state contribution is $g_1\omega_1^{-p}$. Evidently as $p\rightarrow \infty$, $\zeta(p)\rightarrow g_1\omega_1^{-p}$. But as we shall see, already for $p=10$ the approximation is adequate for the purposes of Sec. V. We first consider fields in an Einstein universe of radius $a$.
For the conformal scalar field $\omega_ia=i$, $g_i=i^2$ with $i=1,2,\cdots,$. Therefore $\zeta(p)=a^p\zeta_R(p-2)$, where $\zeta_R$ denotes the Riemann $\zeta$ function. Thus $\zeta(10)=1.004\omega_1^{-10}$. For the electromagnetic field, $\omega_ia=i+1$, $g_i=2i(i+1)$ with $i=1,2,\cdots$. Thus $\zeta(p)=2a^p[\zeta_R(p-2)-\zeta_R(p)]$ and $\zeta(10)=1.05\times 6\omega_1^{-10}$. For the two-component neutrino field $\omega_ia=i+\frac{1}{2}$, $g_i=2i(i+1)$ with $i=1,2,\cdots$. Thus $$\zeta(p)=\frac{1}{2}a^p[(2^p-4)\zeta_R(p-2)-(2^p-1)\zeta_R(p)],$$ so that $\zeta(10)=1.02\times 4\omega_1^{-10}$. Of interest in Sec. V is $\zeta(p)^{1/(1+p)}$. For $p=10$ the fractional errors incurred in this quantity by replacing any of the above $\zeta$ by $g_1\omega_1^{-p}$ are less than $\frac{1}{2}$ percent.
Let us now consider the various fields in a sphere of radius $R$. According to Appendix $A$, for the scalar field $$\zeta(p)=R^p\sum^\infty_{l=0}(2l+1)\sum^\infty_{n=1}j_{n,l}^{-p}.$$ For the neutrino field $\zeta$ is just twice this expression. A good approximation to these $\zeta(p)$ is obtained by summing only over the eigenvalues up to five times larger than $\omega_1=\pi/R$ (about $220\omega$'s). One finds $\zeta(10)\simeq 1.10\omega_1^{-10}$, "for both the scalar and neutrino fields. For the electromagnetic field, $j_{n,l}+j'_{n,l}$ replaces $j_{n,l}$ in formula. A good approximation to this $\zeta(p)$ may be obtained by summing over the eigenvalues up to five times larger than $\omega_1=2.082/R$ (about $140\omega$'s). One finds $\zeta(10)\simeq 3.33\times 3\omega_1^{-10}$. In all these cases the fractional errors incurred in $\zeta(p)^{1/(1+p)}$ by replacing $\zeta(p)$ by $g_1\omega_1^{-p}$ are $1-10$% for $p=10$.