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From several unrelated sources (such as Scott Aaronson's discussion of hypercomputation or this article about a bound on the number of degrees of freedom of any theory with a positive cosmological constant), I have bumped into the claim that the universe contains a finite amount of information (on the order of $10^{122}$ bits), backed by references to the Bekenstein bound.

I followed the references to Bekenstein's original article. The logic seems to be approximately this:

  • General relativity theorems show that (classically) the total area of black hole event horizons cannot decrease.
  • If we assume that black holes have zero entropy, then we can break the second law of thermodynamics by throwing matter with some entropy into a black hole.
  • Therefore, it is natural to extend the concept of entropy by adding a contribution from black holes, whose entropy should be directly proportional to their surface area.
  • The proportionality constant $\frac{S}{A}$, in Planck units, is $\frac{1}{4}$. (This is the step I am having trouble with.)
  • When a black hole absorbs an object of energy $E$ and radius $R$, its area must (for general-relativity reasons) increase by at least $8 \pi ER$. This increases the black hole's entropy by $2 \pi ER$. For the second law not to be violated, the body cannot have more entropy than this. This creates a bound on $\frac{S}{E}$ that we can then plug values for the entire observable universe into, getting the $10^{122}$ bits.

But I am having trouble tracking down where that proportionality constant came from. Bekenstein provides three references for it: Ref. 1, Ref. 3, Ref. 6.

  • In Ref. 3, Bekenstein calculates it from a thought experiment involving dropping a single particle into a black hole. He makes the assumption that the information associated with the particle is at least one bit (as the answer to the question "does the particle exist?"), but he himself considers this a lower bound, with the true value probably not far away, but unspecified.
  • Ref. 6 is also earlier work by Bekenstein; it links to Ref. 3 and makes the additional observation that a higher proportionality constant would forbid particles from being captured by a black hole reversibly (with $\Delta S_g = 0$, where $S_g$ is the generalised entropy, that is, classical entropy plus black hole entropy). This provides some plausibility-based argument for this precise value, but not a derivation from uncontroversial facts.
  • Ref. 1 seems to be the original derivation of Hawking radiation; it's above my skill level and I cannot see whether at any point there is a similar assumption made.

In particular, I could imagine the proportionality constant being bigger. There are many thought experiments considered in these papers and their references that show various processes (which involve dropping various kinds of particles or radiation into a black hole) adhere to the bound, but they would also adhere to the same bound if the proportionality constant were artificially increased tenfold. To disprove this possibility, one would instead want some opposite process, where the black hole's surface area decreases while classical entropy increases. I believe Hawking radiation is of this nature. Is there a better, less plausibility-based assumption in that derivation?

These papers are also rather old, perhaps there have been better treatments since then.

My question is: Does Hawking radiation, or some other thought experiment, provide an upper bound on the entropy of black holes? If so, where in the derivation does one find the upper bound on the amount of information? How is the link between black hole area and bits of information established? Does it rest on a similar assumption of "one particle, one bit"?

Note: The question about the proportionality constant has been asked before here, but I hope you agree my question is far more focused and would not be likely to find answers there.

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    $\begingroup$ Real blackholes, if they exist, have to be more complicated than the idealized featureless blackholes in vacuum. In particular, on can't just get rid of something by throwing it in. Lots of mass thrown its way should hover over the horizon, slowed down, with non-zero entropy. So that entropy does not get "lost from the universe" in any finite time; it just concentrates near the horizon. This hovering entropy cannot be proportional to black hole area as given by the usual formula(proportional to mass squared), because the hovering entropy depends on thermodynamic state of the hovering bodies. $\endgroup$ Commented Sep 29 at 12:17
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    $\begingroup$ I posted a link with references. If you believe Hawking's $S = A/4G$ is correct then a non-black hole with area $A$ must have $S < A/4G$. If it had more, turning it into a black hole by adding mass would violate the second law. And yes, density of states counts the dimension of the Hilbert space, $\log_2$ of which is the number of qubits worth of information. $\endgroup$ Commented Sep 29 at 16:44
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    $\begingroup$ Unfortunately, those heuristic arguments are just that -- heuristics. If you actually want to derive the numeric factors, there's no alternative to going through a proper derivation of Hawking radiation. There are a couple ways to shortcut the calculation (e.g. by making an analogy with the Unruh effet) but no way to remove the math entirely. $\endgroup$
    – knzhou
    Commented Oct 7 at 18:15
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    $\begingroup$ Specifically, the number you need to get is the temperature of Hawking radiation from a black hole, which is $1 / (8 \pi M)$. From that, you can deduce the entropy using basic thermodynamics, without having to resort to any arguments involving "bits" or whatever else. $\endgroup$
    – knzhou
    Commented Oct 7 at 18:18
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    $\begingroup$ Also, because there's always been massive public interest in Hawking radiation, you're not going to find anything good on that subject through an internet search. 99.9% of the results are going to be extremely heuristic, sometimes nonsensical arguments designed for nonphysicists to find vaguely plausible. The real calculation is done right in Hawking's original papers 50 years ago. Some more modern graduate textbooks cover it too, but they essentially give the same argument. There's also a more modern picture involving tunneling, but it's also technical. $\endgroup$
    – knzhou
    Commented Oct 7 at 18:22

2 Answers 2

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An reasonably clear explanation is given in General Relativity and its Applications Black Holes, Compact Stars and Gravitational Waves ,By Valeria Ferrari, Leonardo Gualtieri, Paolo Pani, see chapters 20.3 and 20.4.

In a nutshell:

  1. The fact that the area of a black hole always increases can be compared to how the entropy of a thermodynamic system is always increasing. This lead to conjecture that S is proportional to A for a black hole. So $S = \alpha A$ for an unspecified constant $\alpha$, the analogue of the second law of thermodynamics. Also one can derive the first law of black hole thermodynamic relating $dM$ and $dA$.
  2. Bekenstein imagines dropping a massive particle into a black hole. The minimum amount of entropy increase is going to be $ln2$, that is 1 bit of information. This means that $\alpha \approx ln2 / dA_{min}$.
  3. By the uncertainty principle there is an uncertainty in the position of the particle, with associated proper length of the order of the Compton wavelength. From this one can compute the associated $dM_{min}$, by requiring that the proper distance between the horizon and the point in which the particle is released equates the Compton wavelength.
  4. Now using the first law one can plug $dM_{min}$ and $dA_{min}$ and can find that $\alpha \approx 1 / \hbar$, but the precise factor is still not fixed. One can also compute the associated temperature, in term of $\alpha$.
  5. Hawking performed a calculation in QFT in curved space time to show that a scalar field coupled to the black hole effectively behave as if the black hole is emitting radiation, as a black body of temperature T. From the temperature found one can see that the numeric value of $\alpha$ is 1/4.
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    $\begingroup$ To be picky, it's certainly not true that the minimum change in entropy is 1 bit. In information theory, one considers smaller entropy changes all the time. For example, if a biased coin comes up heads 90% of the time, the entropy associated with its result is $-0.9 \, \text{log}_2 \, 0.9 - 0.1 \, \text{log}_2 \, 0.1 = 0.47$ bits. $\endgroup$
    – knzhou
    Commented Oct 9 at 17:32
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The OP asks I am having trouble tracking down where that proportionality constant came from

You don't need Hawking radiation to make a semi-classical argument why the proportionality constant is $1/4$ rather than something else.

As @Rexcircus above says, we can conjecture that entropy $\propto$ area. The surface area of a sphere: $S=4\pi r^2$. If we consider the smallest semi-classical radius, the Planck length, and use natural units so $L_{Planck}=1$ we have $S=4 \pi$. One unit of entropy for one unit of area. Nice.

PS - In a semi-classical analysis, the Planck length plays the role of a minimal length. But don't take my word for it, you can read this.

However, the Schwarzschild radius of the smallest semi-classical mass (Planck mass $M_P$) black hole (which is a sphere) is: $R=2GM_P/c^2$. In natural units $G=\hbar=c=1$ so $M_{P}=1$ and then $R=2$. Still, this Planck mass black hole must exhibit the minimum semi-classical entropy we just figured out. How to do that? Well, we need a proportionality constant, call it $1/P$. As we can see below, the formula only works when $P=4$, i.e. $1/P=1/4$. Job done.

\begin{equation} S= 4\pi = 4\pi R^2 \frac{1}{P}=\frac{A_H}{4} \end{equation}

PPS. For those who doubt that the Planck mass is the correct mass to use in a semi-classical analysis, read this answer aka the mass at which it happens is about the Planck mass

Bonus: If you want to know where the $S \sim 10^{122}$ of the Universe comes from, it is the entropy (nats) of the cosmic event horizon, which you can read about in this wonderful paper by Lineweaver (Eqn 49). The connection between bits and nats in in my answer here.

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  • $\begingroup$ As to whoever gave this a negative vote - show me the error in this argument. Pfft. $\endgroup$ Commented Oct 13 at 6:14
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    $\begingroup$ I'd guess the phrase "smallest semi-classical radius, the Planck length" is rather problematic and hides a lot of assumptions. This is a reasonable dimensional analysis argument, so no downvote from me, but it doesn't exactly give one an understanding... $\endgroup$
    – Kotlopou
    Commented Oct 13 at 8:39
  • $\begingroup$ @Kotlopu - in a semi-classical analysis, i.e. until a better theory of quantum gravity is devised, the Planck length is the best estimate we have for a minimum length. $\endgroup$ Commented Oct 13 at 11:04

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