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.

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    $\begingroup$ @Rexcirus That's wrong. "The units in which $S = (1/4) A$" are units in which Boltzmann's constant and the Planck length have both been set to $1$. In these units, $S/V$ is also dimensionless. $\endgroup$ – tparker Jul 6 '17 at 21:49
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    $\begingroup$ The fact that you are in a unit system in which the numerical value of the planck length is 1 doesn't mean that a volume and an area have the same dimensions. And of course physical statements cannot depend on the choice of the units that you make. $\endgroup$ – Rexcirus Jul 7 '17 at 9:35
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    $\begingroup$ @Rexcirus I think setting $l_P=1$ doesn't mean setting just its numerical value $1$. It means making it dimensionless. Just like when we set $c=1$ it doesn't mean that only its numerical value is $1$, rather we mean that it is dimensionless and thus, space and time have the same dimensions. $\endgroup$ – Dvij D.C. Jul 7 '17 at 9:46
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    $\begingroup$ @Rexcirus is right. The dimensions are still there, you just don't see them. In the international system for example the meter is the unit of length, so we have $m=1$. Even if we avoid writing the units explicitly, like it is usually done when working with natural units, we still cannot compare surfaces to volumes, because the first have units $m^2$, while the second have units $m^3$. $\endgroup$ – valerio Jul 7 '17 at 11:44
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    $\begingroup$ @valerio92 I think your analogy is not appropriate. In relativity, we can very well add $t$ with $x$ and so on. More relevantly, $v+v^2$ is a valid thing to do in relativity. And all that happens 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. $\endgroup$ – Dvij D.C. Jul 7 '17 at 11:53

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)

  • $\begingroup$ I was relying on this paper for assuming that the volume of a black hole would be proportional to $r^3$ for at least $3+1$ $D$ Schwarzschild black hole: arxiv.org/abs/hep-th/0508108 $\endgroup$ – Dvij D.C. Jul 7 '17 at 3:47
  • $\begingroup$ @Dvij Even in this case, there is still a technical problem (see updated answer). $\endgroup$ – valerio Jul 7 '17 at 7:05
  • $\begingroup$ I completely agree with your arguments that we always have the entropy-area relationship. I have no doubt/argument about it. On the contrary, I use that fact that entropy is proportional to the area to formulate my doubt about holography (i.e. "about where the information really lives" hoping that it is a valid concern). So, no arguments against your answer where you explain why the entropy should be equal to the area by four. Now, I also agree that the volume of a black hole is a tricky concept (otherwise there wouldn't be a paper titled "The Volume of a Black Hole" ;-))... $\endgroup$ – Dvij D.C. Jul 7 '17 at 8:56
  • $\begingroup$ But, assuming that the paper I attached has some merit based on its peer-reviewed publication and not-so-bad peer-reviewed citations, it is a correct statement to make that the entropy density is more below a certain value of mass if we calculate the density volumetrically as compared to if we calculate it areally. And this claim suffers no troubles of dimensional inconsistencies if I use the units in which $l_p=1$. Whether this claim has anything interesting to say or it is just a mathematical weirdness with no real physical excitement whatsoever is something that I still wonder. $\endgroup$ – Dvij D.C. Jul 7 '17 at 9:00

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.

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    $\begingroup$ Oh, great. I thought of doing the explicit calculation but never bothered to do so actually! Thanks for a clear and straight-to-the-point answer. $\endgroup$ – Dvij D.C. Jul 7 '17 at 3:45
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    $\begingroup$ This reasoning is circular, you are assuming what do you want to prove. 1)That the only natural scale it's the Planck scale. It's reasonable, but again, it's circular. Many proposal of quantum gravity forecast effects before this scale (fuzzballs, firewalls, ...). 2) That the constant of proportionality k is approx. 1. Why it cannot be 10^80 ? $\endgroup$ – Rexcirus Jul 7 '17 at 9:23
  • $\begingroup$ @tparker, By the way, I didn't mean to ask whether $S$ would start becoming proportional to $V$ or not. I was only wondering that $S/V>S/A$ would become an interesting thing for the question of "where the information lives" or not. Anyway, your answer gives a direction to think for the same and the result still remains the same that $S/V>S/A$ would happen when $r\sim 1$ in the Planck units (with $l_P =1$). $\endgroup$ – Dvij D.C. Jul 7 '17 at 9:51
  • $\begingroup$ @Rexcirus I don't think anything is circular here. Let's say we have, $S=\pi r^2$, $S/A=1/4$ and $V=4 \pi r^3/3$. Therefore, $S/V=1/4$ only when $r=3/4$. I am working with $l_P=1$ - so you can say $r=3l_P/4$. Clearly, the transition is happening around the Planck scale. $\endgroup$ – Dvij D.C. Jul 7 '17 at 12:07
  • $\begingroup$ @Rexcirus The two assumptions that you list are actually two ways of phrasing the same assumption. If $k \sim 10^{80}$, then the Planck scale is not the natural scale to use. "Natural scale" means that important numbers are $o(1)$ in those units. Also, whether justified or not, it's extremely common to assume that most interesting quantum gravity effects kick in around the Planck scale. That's why the hierarchy problem and the cosmological constant problem are considered "problems" - because they seem to violate that assumption. $\endgroup$ – tparker Jul 7 '17 at 14:11

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