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The Bekenstein bound is a limit to the amount of entropy a thermodynamical system can have. The bound is given by the following expression: \begin{equation} S \leq \frac{2 \pi k R E}{\hbar c} \end{equation} where $k$ is Boltzmann's constant, $R$ is the radius of a sphere that can enclose the given system, $E$ is the total mass–energy including any rest masses, $ħ$ is the reduced Planck constant and $c$ is the speed of light.

The equality is reached for Black Holes.

Now, a system is in thermodynamical equilibrium when the entropy of the system is in a maximum and the constrains of the system( like pressure, volume, etc.) are satisfied. In our daily live, when we consider thermodynamical systems the bound is never achieve; only thermodynamical systems at the scale of astronomical objects seem to satisfy it.

Why is the equality only achieved at certain scales?

yess
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2 Answers2

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The Bekenstein bound tells the maximum entropy that can be contained within a given volume. It does not tell the maximum entropy of a closed system with a fixed energy density. The distinction is very important here. For a system with a relatively low energy density, the maximum entropy state is not a black hole. It is, instead, diffuse radiation. This is why Hawking Radiation exists. When a black hole radiates into empty space, total entropy is increased (1).

So, that's part of the answer to your question- low energy thermodynamic systems don't settle into black holes because that's not their highest entropy state in the first place. Your followup question might be: okay, so why is the limit for entropy density only satisfied on astronomical scales?

I think the best answer I can give you is this: gravity is unique among the forces in that it is always attractive, and as a result it has much more capability to put matter, and thus entropy, in a compact area. But gravity is also, famously, much much weaker than any of the other forces, so it is only relevant on large scales where all the other forces are cancelled out due to screening effects. There is almost certainly a good reason that gravity has these two distinctive properties, but as far as I know we will need an understanding of quantum gravity to really address those questions.

Rococo
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The volumetric scale of a black hole has nothing to do with its properties.

Black holes exhibit so many downright strange and extreme behaviors not because of their size, but because of the ridiculous amount of mass they have.

Think about newton's law of gravitation for a moment. The gravitational force between two objects is inversely proportional to the square of the distance between them and proportional to both objects' masses.

Now imagine a spherical star. The gravitational force felt by another object from the star is directed towards the very center of the star, almost as if the entire mass of the star is localized at the center point and the volume of the star simply represents a physical boundary that keeps other matter from coming close to the center.

If you shrunk the star down to a size that is point-like in comparison to its original volume but kept all the mass (like how stars collapse to form black holes), you'd be able to achieve a much higher gravitational force by newtons equation because you'd be able to come much, much nearer to the spatially localized source of the gravitational force.

The same concept applies on any scale. If you somehow squeezed the mass of an entire continent into a volume the size of a pencil tip (or something like that), there might be an event horizon around the pencil tip that represents the distance from the center where gravity becomes too intense even for light to escape. Essentially, you'd have a black hole.

Rick Sanchez
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