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The entropy of a non-rotating, non-charged black hole is only proportional to the surface area of its event horizon (Bekenstein, 1973).

The area of the event horizon of a spherical black hole is only dependent on its radius (equation for surface area of a sphere), which in turn is only dependent on the mass of the black hole (Schwarzschild radius).

All in all, we can take the entropy of a black hole with these characteristics as a function of its mass.

Now let us consider an object falling into a black hole (with these characteristics), which then decreases the entropy of the whole universe minus the black hole. This lost entropy can then be found in an increase of entropy of the black hole, which must depent on mass alone.

Because we can throw every object into a black hole, this means that the entropy of the entire universe depends on mass alone, which I don't think is logical as entropy increases with a different order of the same mass.

What went wrong here? Either there exist no black hole with these characteristics, or there is some mistake in my reasoning.

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Now let us consider an object falling into a black hole (with these characteristics), which then decreases the entropy of the whole universe minus the black hole. This lost entropy can then be found in an increase of entropy of the black hole, which must depent on mass alone.

When 1 kg of extremely high entropy matter is lowered gently, no new entropy is generated.

When 1 kg of extremely low entropy matter is lowered gently, no new entropy is generated.

When 1 kg of extremely high entropy matter is dropped, lot of new entropy is generated.

When 1 kg of extremely low entropy matter is dropped, even more new entropy is generated, than in the previous case.

(When 1 kg of matter is lowered gently, mass of the black hole might increase by one milligram)

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  • $\begingroup$ “When 1 kg of matter is lowered gently, mass of the black hole might increase by one milligram” - This is actually correct +1. The mass of the black hole would increase by 1 kg only when a 1-kg object starts falling from rest at infinity. The closer to the black hole it starts falling from rest, the less extra mass the black hole gets. The specific value is defined by the time dilation. $\endgroup$
    – safesphere
    Commented Nov 7, 2023 at 3:42
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Your question is valid and your deductive reasoning is plausible. The entropy of the universe indeed increases as mass is thrown into the black hole. It is so far valid. However, it is invalid to mention that entropy is dependent on the mass alone which is supported by the equation mentioned:

2GM/c^2

The problem arises in the type of mass that has entered the black hole. If we remember the Hawking Radiation, by which black holes may lose matter(which means an increase in entropy), we realise that indeed, it is not solely dependent on matter. For instance, if a particular antiparticle enters the black hole, then the 'antiparticle's particle' would be radiated into the universe, increasing the entropy.

The point being made, is that the universe is not solely dependent on its mass, because of the type of mass entering. Yes, if ordinary matter enters a black hole, the entropy increases; nevertheless, we may have antimatter entering the universe as well.

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  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Nov 7, 2023 at 7:15
  • $\begingroup$ Thank you very much, this makes a lot of sense. Should I intepret this as: antimatter decreases entropy? How does order and symmetry play a role here? Is there a geometric meaning to antimatter and matter? $\endgroup$
    – Koen de Jong
    Commented Nov 7, 2023 at 8:32
  • $\begingroup$ Another point to be made: when does antimatter come in the equations? According to your reasoning (which I accept), either at the Bekenstein equation for entropy, or at the equation of a sphere, or at the Schwarzschild radius, antimatter come in. You mention that the Schwardzschild radius equation includes the physics of antimatter, but which term represents this? $\endgroup$
    – Koen de Jong
    Commented Nov 7, 2023 at 15:07
  • $\begingroup$ @safesphere, I accept all of your statements, except the one on the geometric meaning of matter and anti-matter. What I meant by that was the fact that with the same mass, an object can have a different entropy if we alter the geometric configuration. Thus, if solely the nature of matter (antimatter or matter) serve as solutions for the problem that I state, how does geometry play a role? $\endgroup$
    – Koen de Jong
    Commented Nov 7, 2023 at 19:32
  • $\begingroup$ @safesphere, I have seen your comment on my original post about maximum entropy, but how does this explain the problem? I consider a mass falling into a black hole. The black hole entropy increases by some mass (according to my reasoning). Then we reason backwards that this increase of entropy of a black hole equals the decrease of entropy of the rest of the universe. Thus, if we write the decreased entropy (which we can chose to be all entropy, given that everything can fall in a black hole, theoretically) in terms of mass, we can write the original entropy of any object in terms of mass. $\endgroup$
    – Koen de Jong
    Commented Nov 7, 2023 at 19:36

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