Is entropy high or low in a black hole?

Question

Just wondering. I guess you can argue either way depending on the arguments you use.

Answer 1

It is high, but only by convention.

In a purely physical sense, black holes have zero entropy, since they have no hair. The mass, charge, rotation, and motion is all that is knowable about them, and all these quantities are accessible to the observer. So there is only one microstate corresponding to its macrostate, so the entropy should be ln(1) = 0.

However, this Bothers Us because it allows any entropy that you throw into a black hole to seemingly disappear. The 2nd Law survives, however, because that entropy is merely stored, not lost. When the black hole evaporates into subatomic particles, those particles will have enormous entropy. Therefore, we cheat a little bit, and we change the definition of entropy from the log of the number of microstates that the black hole has to the log of the number of microstates that its evaporation products *will have* when it evaporates, which, as has been said, happens to be proportional to the surface area of the event horizon.

Yes, it is lovely to be able to make up your own laws of physics, since, after all, that is what I get paid to do for a living.

Answer 2

The entropy of a black hole is one fourth of the area of the event horizon, so the entropy gets smaller and smaller as the black hole decays and the event horizon area becomes smaller and smaller.  

Answer 3

It's sad to see that some people like Kelly and Jose Frink can make up there own Laws of Physics without researching anything that Stephen Hawking and other Astrophysicists have proven.

Answer 4

I kinda figured it would be high (while what ever item is being drawn into it), since supposedly a black hole sucks up everything in its event horizon (even light) and crushes it/rips it apart until only atoms are left. But then nothing can change so there wouldn't be any.

Although since black holes are assumed to have zero temperature it was also assumed that they had zero entropy. But since zero temperature implies infinite changes in entropy with any addition of heat (anything that the black hole sucks up that has any sort of heat to it), implies infinite entropy.

Hope it helps at least a little.

Answer 5

Black Hole entropy is proportional to its area.

The only way to satisfy the second law is to admit that the black holes have entropy whose increase more than compensates for the decrease of the entropy carried by the object that was swallowed. The black hole entropy is also the maximal entropy that can be squeezed within a fixed volume.

The entropy of a black hole is expressed as

S = Akc^3 / 4hG

where A is the area of the black hole, k the Boltzman's constant, h the Planck's constant, c the speed of light and G the gravitational constant.

Edit-Jose Frink; please read the source which I have cited: In Regarding Hawking Radiation:

"With the Hawking radiation, there is a decreasing of the area of the black hole, due to the decreasing of its mass with the evaporation. We've seen that this area is comparable to the entropy, but as the loss of entropy of the black hole is exactly compensated with the increase of entropy of the thermal radiation, there is no violation of the second principle of thermodynamics"