Where does entropy of whatever collapses into a black hole disappear?

Question

A star before it implodes is a ball of hot gas,
its entropy is huge. After implosion all we have
is mass, angular momentum, and electiric charge.

But according to the 2nd law of
thermodynamics entropy cannot decrease.

Answer 1

Mass, angular momentum, and electric charge being the only properties does not preclude other properties that are derivable from these such as, to pick a trivial example, charge to mass ratio. Energy (mc^2) and temperature are two other such properties and, with it, entropy (temperature is the derivative of energy wrt entropy). Hawking proved that black holes emit a black body spectrum (aka Hawking radiation) characteristic of a well defined temperature and, therefore, do indeed have entropy - many orders of magnitude more than the original star, in fact.

The radiation is emitted just exterior to the event horizon where space is highly distorted to the point where virtual particles tunnel out of the vacuum to become real due to the gravitational field intensity. The excited vacuum, therefore, is responsible for the great multiplicity of quantum states whose logarithm defines entropy.

Answer 2

dude, first of all, our sun will not be a black hole. look Stars like the Sun just aren't massive enough to become black holes. Instead, in several billion years, the Sun will cast off its outer layers, and its core will form a white dwarf - a dense ball of carbon and oxygen that no longer produces nuclear energy, but that shines because it is very hot. A typical white dwarf is about as massive as the Sun, but only as big as Earth, which is one percent of the Sun's present diameter. and also for more info, look this The Sun is way too small to become a black hole. Stars have to be more than double the size of the Sun to become a black hole. When the Sun runs out of "fuel" (hydrogen), the collapse will increase the pressure in the core, which will kick start the fusion of Helium. However, this will not last very long: fusing Helium provides a lot less total energy than fusing hydrogen. But it provides it faster: the Sun will bloat out and become a red giant. At some point in the process, the Sun will shed out a large portion of its envelope, leaving only the core. At that point, there will be insufficient pressure to generate any more fusion. The remaining core will continue to glow (for another 5 to 10 billion years) just from the accumulated heat. That is the white dwarf stage. Because there is no fusion at the centre to push against the pressure of gravity, the matter in the core will become "degenerate" (the electronic shell of each atom will be snug against the electronic shell of the next atoms), making the matter extremely dense. The good old description is something like: a teaspoon will weigh many tons. But not a black hole. Black holes occur when much more massive stars collapse: the electronic shell is insufficient to stop the collapse and the collapse continues, until the nucleus of each atom is one against the other, and even that is insufficient. The collapse then continues forever, with nothing stopping it. Our Sun is way too small for that. Edit: oh, but for the first part, impossible. second part, impossible third part, don't even think about it.

Answer 3

Actually a star's entropy is not that high, because the energy it contains is confined to a relatively small space (on a unversal scale) and hence a relatively small number of microstates.

But in answer to the main thrust of your question, a black hole has the highest possible entropy per unit volume. Bekenstein showed that the entropy of a black hole is proportional to the area of its event horizon.

Answer 4

Ask Stephen Hawking and Kip Thorne