# The Problem

I compute: $$I \leqslant 4.6\times10^7 bits/eV m \cdot R \cdot E$$

I chose these units because they are convenient for dealing with photons. In particular, I wish to consider a 1 m sphere containing a single 3 MHz photon, which has an energy of $$1.24\times10^{-8} eV$$. This gives the Bekenstein bound of $$I <= 0.57 bits$$.

However, we know that photon has spin angular momentum, which requires at least 1 full bit to describe. Therefore, it would seem that this space is not sufficient to contain the photon? According to Bekenstein, we need at least a 2 m sphere to hold this photon?

# Possible Resolutions

Answers here suggest that the Bekenstein bound does not extend to the quantum realm, and should only be applied at macroscopic levels relevant to GR. Ok. Then let's increase the number of photons by a factor of $$2\times10^{27}$$. Now we have a Joule's worth of photons, more or less, all meeting within a 1 m sphere. This should be adequately macroscopic, no? But notice the linearity: while we increase the energy by $$2\times10^{27}$$, giving us many, many bits, we also increase the number of photons by the same amount, correspondingly increasing the "demand" for bits. Each of those photons has its own spin angular momentum, and there is no a priori reason for them to be correlated. In fact, it should be possible to create a cloud of such photons with every combination of SAMs. Therefore, we need $$2\times10^{27}$$ bits, but Bekenstein only gives us $$1.14\times10^{27}$$.

The only factor I could imagine is the fact that our 3 MHz photons have a wavelength of about 100 m. Perhaps the fact that the photons don't "fit" inside a 1 m sphere is the source of the problem? But, of course, there should be no problem building, say, a 3 MHz 100 W transmission antenna and pointing it into a 1 m spherical space, and getting more than enough of the relevant photons within that space, right?

What am I missing?

• $I <= 4.6\times10^7 bits/eV m$ Did you forget the $R$ and the $E$? – G. Smith Oct 6 '20 at 22:59
• That's what the eVm are in the denominator. It's normalized to 1 m and 1 eV. Multiply by R and E as desired. – Lawnmower Man Oct 6 '20 at 23:02
• That may be what you intend, but as written it makes no sense because the units of $I$ are bits, not bits per eV per meter. Look at the third formula in the Wikipedia article. – G. Smith Oct 6 '20 at 23:10
• Wikipedia gives the formula where I is stated in terms of bits, but it is just as legitimate to specify I as a "bit density" with units bits/eVm. Even so, I changed it to bits to eliminate any further confusion. – Lawnmower Man Oct 6 '20 at 23:20
• It’s not legitimate to use $I$ to mean two different things in the same question. Your inequality now makes sense. – G. Smith Oct 7 '20 at 1:53

First, Bekenstein's bound has been found to hold in all physically-sensible relativistic quantum field theories for which it has been checked. See ref 1 for an example. It definitely is applicable to the quantum realm.

Second, Bekenstein's bound is a bound on the number of mutually orthogonal quantum states that have volume $$\leq \frac{4\pi}{3}R^3$$ and energy $$\leq E$$. We can also express this in terms of the number of bits of information that can be stored in that volume with that energy, but it's the same idea either way. It's the number of different ways the volume can be populated with physical things, if the energy is not allowed to exceed $$E$$.

Examples:

• If $$E=hf$$ with $$f=3$$ MHz, then nothing with energy $$\leq E$$ can fit in a sphere of radius $$R$$. In particular, a photon with energy $$\leq E$$ has a wavelength $$\gg R$$, so it doesn't fit. Neither does anything else with so little energy.

• If we increase the upper limit on the energy to, say, $$E=100$$ Joules, then Bekenstein's bound says that lots of mutually orthogonal quantum states have $$\leq$$ that much volume and $$\leq$$ that much energy.

A photon is a particle in the sense that photons can be counted, but a photon is no more pointlike than a classical EM wavepacket with the same wavelength. An optical photon is pointlike when compared to a human eye, but it's not pointlike at all when compared to an atom, and a $$3$$ MHz photon is not even as pointlike as a bus.

We can build a $$3$$ MHz $$100$$ Watt transmission antenna and point it into a $$1$$ meter spherical space, but radio waves with wavelengths $$\geq 100$$ meters can't produce features of size $$\sim 1$$ meter. If we do something to trap the EM radiation inside the sphere, then Bekenstein's bound tells us that we are no longer dealing with photons having energy $$hf$$ with $$f=3$$ MHz. In other words, if we actually solve this boundary-value problem for the EM field to see what cavity-modes are possible, we will discover that the minimum possible energy in one quantum of such a mode is $$\gg h\times 3$$ MHz.

Here's an excerpt from a review by Bekenstein himself (ref 2):

it is not always obvious in a particular example how the system avoids having too many states for given energy, and hence violating the bound. We analyze in detail several purported counterexamples of this type (involving systems made of massive particles, systems at low temperature, systems with high degeneracy of the lowest excited states, systems with degenerate ground states, or involving a particle spectrum with proliferation of nearly massless species), and exhibit in each case the mechanism behind the bound's efficacy.

1. Schiffer and Bekenstein (1989), "Proof of the quantum bound on specific entropy for free fields," Phys. Rev. D 39, 1109, https://journals.aps.org/prd/abstract/10.1103/PhysRevD.39.1109

2. Bekenstein (2005), "How does the entropy/information bound work?", https://arxiv.org/abs/quant-ph/0404042

• To close a loophole here, what if we don't trap the 3 Mhz photon in a cavity, but just say that we know at time <1m/c after the antenna emits it, it has to be within a 1m sphere around the antenna? For 1 photon, this fails because we can't identify the emission time with precision. But if we emit lots of photons, then it seems at any given time there ought to be at least 1 photon propagating (or perhaps coalescing) in the 1m sphere. But now I'd say the issue is we have to consider the antenna itself and its mass and oscillations, which changes the inequality. Does that sound right? – Felix Wotter Oct 7 '20 at 3:23
• Great answer! If we pick E = 1.2 μeV, R = 1 m, then we should have about 57 bits available. This is just enough space and energy to fit a single 300 MHz photon. How should we interpret the meaning of those 57 bits? – Lawnmower Man Oct 7 '20 at 5:50
• @FelixWotter You're right: in that case, when checking the bound, we would have to include the energy of the antenna itself, or whatever part of it is inside the sphere of interest. That will be a lot of energy, because it must include the antenna's mass-energy ($mc^2$). The $E$ in the Bekenstein bound is the energy of the gravitating system (even though we're ignoring its gravity), which means we need to include rest-energy. – Chiral Anomaly Oct 7 '20 at 12:39
• @LawnmowerMan In a case where the bound gives $57$ bits, it means that we can find no more than $2^{57}$ mutually orthogonal quantum states, each of which has no more than the specified energy and no more than the specified volume. Just like in statistical mechanics, it's about counting microstates: how many distinct (more precisely, mutually orthogonal) states are compatible with the given macroscopic conditions? "Mutually orthogonal" is important because it prevents us from counting slightly different configurations as distinct states, both in stat-mech and in the Bekenstein bound. – Chiral Anomaly Oct 7 '20 at 12:40
• Ok, but besides a low-energy photon sitting in the sphere, what other quantum states are possible? All other bosons are pretty massive, as are quarks, to say nothing of baryons. Are there really $2^{57}$ ways to pack a single photon into a 1 m sphere? Could you describe a few of them? – Lawnmower Man Oct 8 '20 at 0:08

If we have photons with different spins, we can paint a picture using those different photons as pixels.

If the photons are for example 1m long, 1m wide and 1m deep, then the picture can't be very sharp.

If we stack 8 photons at the same place, then the pixel at that point can have one color out of 256 possible colors.

We would have one byte of information in one cubic meter, if pixel's volume was one cubic meter, and if we put eight photons into a pixel. Or should I say voxel.

Oh, the 256 is a wrong number, because the photons are not in any order in the voxel.

Anyway, the entropy in this case is proportional to the logarithm of the number of all possible pictures. And the number of all possible pictures is proportional to the energy of the photons.