More (Dense) Information in Bulk than on Surface?

Question

The entropy of a Schwarzschild black hole is proportional to $m^2$ where $m$ is the mass of the black hole. The volume of the black hole would be proportional to $m^3$ and the area would be proportional to $m^2$. Thus, with an increasing mass, the ratio of the entropy to volume would keep on decreasing but the ratio of the entropy to the area remains constant - always.

If we consider the entropy to be a direct measure of information then information per unit volume keeps on decreasing with increasing information but the information per unit area remains constant.

In my very limited knowledge about holography, I think this observation is one of the basic ideas behind holography: although we can have a large volume, we don't have enough information inside it if we expect every unit of volume to have some information on its own. Rather, the information seems to live on the surface where the larger the area, the larger the entropy (and in the same proportion).

But below a certain value of mass, the ratio of entropy to volume would become greater than the ratio of entropy to area. I am not sure why but this seems weird in some sense. I understand that the ratio of entropy to area is still constant but if the information really lives on the surface then the fact that the information is more dense in the bulk than it is on the surface seems awkward.

Is this a legitimate concern or there is nothing awkward going on here?

Edit Owing to some discussion in the comments, I would like to clarify that I don't think that raising the issue that area and volume have different units has any curcial relevance here. I work in a system where $l_P=1$. Just like in relativity we can very well add $t$ to $x$ and so on by setting $c=1$, we can compare $A$ and $V$ as the same dimensional quantities by setting $l_P=1$. In relativity $x/t$ is dimensionless owing to setting $c=1$ - not just setting $c$ as the reference quantity for speed but setting $c$ to $1$ - a dimensionless constant. Similarly, if we set $l_P=1$, we can very well have $A/V$ dimensionless.

Answer 1

The entropy of a Schwarzschild black hole is proportional to $m^2$ where $m$ is the mass of the black hole. The volume of the black hole would be proportional to $m^3$ and the area would be proportional to $m^2$.

I think that there are some problems here.

First, the entropy of any black hole (not just a Schwarzschild black hole) is exactly proportional to its area, i.e. the area of the event horizon:

$$S_{BH} = \frac{k_B A}{4 l_p^2}$$

This exact formula links the entropy to the area, and not to the mass. Then, for a Schwarzschild black hole, you have

$$A = 4 \pi r_h^2 =16 \pi \left(\frac{G M}{c^2} \right)^2$$

So that indeed in this case we have

$$S_{BH} \propto M^2$$

For the volume, things are a little more complicated. Indeed, one would be tempted to say that the volume of a Schwarzschild black hole is

$$V= \frac 4 3 \pi r_h^3 = \frac {32} 3 \pi \left(\frac{G M}{c^2} \right)^3$$

However, this would be wrong. The problem is that there is no unique volume that we can assign to a black hole (see for example here or here or here). In a certain way, we can say that the "volume" of the black hole depends on how long the life of the black hole is: for an hypothetical eternal black hole, this "volume" would be infinite. Therefore, I feel like this paradox is not very well defined, because it is not clear what you mean when you talk about the "volume" of the black hole.

Apart from the fact that the area of the black hole, $A$, is a well-defined quantity (unlike the volume), there are very good reasons to believe that the entropy should be proportional to the area, one being the area theorem: the event horizon area of a black hole cannot decrease. This "never decreasing" behavior is very reminiscent of the "usual" thermodynamic entropy, and makes us speculate that $S_{BH}$ must be a monotonically increasing function of $A$ (it turns then out that it is the simplest such function, i.e. a linear function).

But let us assume that we have some good definition of the volume of a black hole $V$, and that $V \propto r_h^3 \propto M^3$. Then, we would still have the problem the entropy per unit volume and the entropy per unit area would have different dimensions. So when you say

But below a certain value of mass, the ratio of entropy to volume would become greater than the ratio of entropy to area.

it is still not clear what you mean, because you would be comparing physical quantities with different dimensions.

See also: Bekenstein-Hawking entropy (Scholarpedia)

Answer 2

Consider a hypothetical black-hole-like object whose entropy scaled with its volume instead of its surface area. Call it a "$V$-hole" to distinguish it from a normal black hole (which I suppose consistency would demand that we call an "A-hole" - I promise I didn't realize that until after I came up with the name "$V$-hole."). Then $S(V\text{-hole}) = n V$ for some constant volume density $n$. $S(A\text{-hole}) = A/(4 l_P^2)$, where $l_P$ is the Planck length, so $R_* = 3/(4 l_P^2 n)$ is the critical black hole radius below which a $V$-hole of radius $R$ has less entropy than an $A$-hole of the same radius.

The only natural scale to set the entropy density $n$ is the Planck length, so we expect $n = k/l_P^3$ for some dimensionless constant $k \sim 1$. The critical radius $R_*$ is therefore $3/(4k)\, l_P \sim l_P$.

But it is believed that black holes below some critical radius around the Planck length cannot exist at all, as explained here. So any physically realizable $A$-hole presumably has less entropy than a hypothetical $V$-hole of the same radius, and the reverse situation that you describe can't actually occur.