# Why aren't we surrounded by Black holes?

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?

• A very simple answer is : because all those constants have been measured to have values such that astronomical sizes are needed for the bound to be fulfilled. Why did we find such values? well the answer becomes the anthropocentric one : we would not be here otherwise. Physics does not answer why really, just how one step leads to the others in our theoretical models based one principles and postulates. May 25, 2014 at 2:56

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.

• Thank you, your answer was really helpful. Would you mind explaining me a little bit why for a system with a relatively low energy density, the maximum entropy state is not a black hole?
– yess
Jun 15, 2014 at 16:45
• @yess: Entropy can also be thought of as a measure of our ignorance. Putting a lot of energy in one place, in an empty space, is going to be simpler, require less data, than that energy spread out as much as possible in the area - ie diffuse photons, the smallest energy unit. Jun 4, 2021 at 21:05
• @CriglCragl That is basically right. I wrote a more recent answer: physics.stackexchange.com/questions/577631/… which goes into much more detail about this problem Jun 5, 2021 at 12:58

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.

• Thanks for your answer. I think what I don't understand is why effectively we don't see process in nature that squeeze continents into the sizes of the pencils. The squeezing only takes place at astronomical scales. Is it related to the weak strength of gravity compared to the other forces?
– yess
Jun 15, 2014 at 16:39
• Yes. If you go look at newtons equation for gravity and compare it to coulomb's law for forces between charges, you'll see that the equations are in nearly the exact same form except they depend on different quantities. What you really want to note, though, is the vast difference between the gravitational constant in newtons equation and the constant "k" in coulombs's equation. This basically results in coulombic forces being much stronger compared to gravitional forces Jun 17, 2014 at 1:44