Does information always gravitate?

Question

I'm trying to wrap my head around Bekenstein's loose argument that a bit of information added to the black hole corresponds to an added Planck surface area to its horizon.

In it, he argues that one can add a bit of information to the black hole by emiting a low frequency photon (of wavelength of the order of the black hole's radius). He seems to reason that this photon contains 1 bit of information because it's location is not well-defined on the black hole, and as such the photon either is inside the black hole, or it isn't. Then he uses the energy of this photon to deduce the change in surface area.

The implication here seems to be that information can be encoded in things that have energy, and energy gravitates, hence we can associate this gravitation to the bit and say that the bit makes the black hole gravitate more.

My question is then two-fold.

Firstly, surely this correspondence (between information and energy, and thus between 1 bit and a Planck area) is not unique? Perhaps naively, I can think of encoding 1 bit of information in other particles, by the same process, for example, an electron of similar wavelength, which will have different energy but same information and as such will gravitate differently. If the spinor nature of the electron doesn't allow it to only carry 1 bit in this way (which I suspect shouldn't be a problem since a photon also has spin, but it seems the information about its spin is not taken into consideration) then how about a Higgs boson of Schwarzschild wavelength? Surely all these objects gravitate differently even though they all have 1 bit of information. Then aren't sentences like "adding 1 bit of information to a black hole adds 1 Planck area to its surface area" a bit arbitrary, since by repeating the same argument with a Higgs boson of Schwarzschild wavelength I would arrive at a different increment in area?

Secondly, if the correspondence between information and its energy isn't unique, is it then possible to add information to the black hole while making it gravitate the same, i.e. is it possible to encode information in a system of zero energy and then feed it to the black hole?

My first guess was no. To encode information on a system would mean to change its state, and to do so I would have to transfer some energy to it. But is it perhaps possible to perform a sequence of operations on a system, each of them adding and subtracting energy from it, such that the net balance is zero, and resulting in 1 bit of information encoded in it? Or does the fact that I performed more than 1 operation already imply that I engraved more than 1 bit?

Its this interplay between information and energy that I'd like to shed some light on.

Answer 1

There's a few things at play here. The increase in entropy you're mentioning is related to the Bekenstein bound $$S \leq \frac{2\pi RE}{\hbar} $$ where S is the entropy of a given region of space, R is the length scale of that region, and E is the energy inside it. But notice something interesting about this bound: Newtons constant doesn't enter into it at all! So even in a world without gravity, Bekenstein's bound still applies. In fact, Casini proved this bound explicitly in vanilla quantum field theory, no gravity at all.

Now, black holes are special beasts because they saturate the above inequality. The first law of black hole thermodynamics says that for a Schwarzschild black hole, $$ dE = \frac{\hbar\kappa}{2\pi} d \left( \frac{A}{4G_N \hbar}\right) =TdS $$ where $\kappa$ is the surface gravity and A is the area of the black hole. This is just a theorem from GR. With Hawking's calculation that black holes have a temperature $T_H = \frac{\hbar\kappa}{2\pi}$, that means that black holes have an entropy $S= \frac{A}{4G_N} + S_0$, for some constant $S_0$. But using the fact that in 4d $R_S = 2G_N M$, you can see that $$ \frac{A}{4G_N \hbar} = \frac{4\pi R_S^2}{4G_N \hbar} =\frac{2\pi R_S M}{\hbar} $$ so we must have $S_0 \leq 0$ by Bekenstein's bound. Considering that $S_0$ is constant, we can always take $M$ small enough that $S$ is negative unless $S_0=0$, which is necessary for the entropy to be a count of microstates $S=\ln \Omega$. Notice that at no point did I use the quantum nature of gravity. We simply used the Bekenstein bound (true of the quantum fields in the semiclassical background of the black hole geometry) and the first law (which is again a theorem of classical GR). Now that we have shown that black holes have an entropy proportional to their area (the constant of proportionality is the inverse Planck area), your claim follows easily. Increasing the total information within the black hole by throwing in a single bit has to increase the area by one in Planck units.

Now I'll offer an argument that this is okay quantum mechanically. In the absence of fundamental symmetries, every state has a unique energy, i.e., the Hamiltonian is non-degenerate. Therefore, the EXACT area of the black hole should tell you EXACTLY what the state is. How many states will have approximately that area? $e^{\frac{A}{4G_N}}$ of them! So actually, you would expect that throwing in a Higgs boson vs an electron vs a photon would change the area slightly differently if you could really keep track of the exact horizon area. If you're not allowed to keep track of that level of precision, and only some window of possible areas between $A$ and $A+dA$ (where $dA \gg 4G_N \hbar $) then they will all contribute to the horizon area in approximately the same way, i.e., they will gravitate in the same way. But if you're talking about electrons or photons, you probably should have some UV cutoff which in effect provides this kind of smearing over the areas, so your concerns of this kind aren't too important actually.

At the end of the day, $dE=TdS$ is the relationship between energy and information that's relevant here. Black holes have a nonzero temperature, so that links the two quantities in the usual thermodynamic way.

Answer 2

If we have a photon and black hole with the same size, the photon can not be lowered to the hole. So all the energy must go to the hole with the photon.

If we have a book and a black hole, the book can be lowered to the hole. So the energy that can not be extracted from the book is proportional to the change of mass of the hole that is required to increase the area of the hole so that the information in the book fits into that area.

Oh yes, if we have a small photon and a big black hole, and we lower the photon into the hole, during the lowering process the photon gets bigger until it's the size of the hole, and additional lowering becomes impossible.

Edit: I mean, what gets bigger, is photon's size at infinity, which is a similar thing as photon's energy at infinity. (Actually, I thought the photon would actually get bigger, but let's forget that)

So that is how different kinds of photons cause the same effect when lowered to the hole. (Lowered, not thrown!)

As for spin, information about spin of a particle that was lowered into the hole can be read from the spin of the hole, so the information did not go into the hole.

As for a case where lowering a particle causes the hole to stop spinnig, the information of the spin of the particle is coded in the amount of energy extracted when lowering the particle.