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Both mass and entropy behave differently for black holes than for normal matter.

For simple Schwarzschild black holes, mass is proportional to their radius. The Bekenstein-Hawing entropy is proportional to their surface.

Of course, the calculations for the two results can be found in the textbooks.

Is there a simple explanation for why the dependencies on the radius differ?

For example, fourfold surface implies fourfold entropy - but only twice the mass. In contrast, in normal matter, fourfold entropy implies fourfold mass.

What is the best way to explain that in black holes, entropy increases more than mass? Somehow, mass and entropy are decoupled.

What is the best way to explain this?

KlausK
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Watch carefully as I wave my hands:

My way naive way of looking at this is to think of the entropy of a black hole as residing at its event horizon- that in some way a black hole is an engine that separates out entropy from mass and stores them separately- at least from a viewpoint outside the EH.

But I have never managed to satisfy myself regarding the question of where the heck a black hole sequesters its mass. If it all fell straight into the singularity, then a black hole would not possess a moment of inertia because the radius of the mass distribution would be zero. So is it stored at the EH surface too? If it were, then the moment of inertia of a black hole would be that of a hollow spherical shell with the radius of the EH, which isn't right either.

I'm very interested in what the experts here can teach me about this.

niels nielsen
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    Isn't the following true? Schwarzchild black holes have no angular momentum. They don't rotate, so not being able to calculate their moments of inertia is unimportant. Rotating black holes have ring singularities, moving the mass away from a central point and making it possible to calculate moments of inertia. – D. Halsey Aug 06 '23 at 17:30
  • @ d. halsey, +1 – niels nielsen Aug 06 '23 at 18:40
  • @D.Halsey The mass of a black hole is outside the horizon. The naive logic based on a flat spacetime is not applicable to strongly curved spacetimes. Your comment is also incorrect for another reason. While Kerr black holes often are called "rotating", the Kerr spacetime is stationary. Such black holes have an intrinsic angular momentum, but nothing actually rotates there. Roughly speaking, this is similar to spin in Quantum Mechanics where particles also an intrinsic angular momentum, but don't actually rotate. – safesphere Aug 07 '23 at 04:29
  • @nielsnielsen "then the moment of inertia of a black hole would be that of a hollow spherical shell with the radius of the EH" - Note that the radial direction from the horizon to the origin is not spacelike, but timelike (not a distance in space, but a period of time). So the spatial radius of the horizon is zero. What you often hear people referring to as the "Schwarzschild radius" is not the radius of the horizon, but the circumference of the horizon divided by $2\pi$ (reduced circumference). The geometry of curved spaces can be hard to visualize: https://i.stack.imgur.com/yM2Md.png – safesphere Aug 07 '23 at 05:24
  • @safesphere I assume you are adopting the point of view of distant observers who observe that nothing ever actually falls into a black hole and stars never completely collapse to black holes. If so, don't you have to accept that the star's rotation before the beginning to collapse is still there? – D. Halsey Aug 07 '23 at 17:39
  • @D.Halsey I see your point. Does the rotation continue forever or does it effectively stop due to the infinite time dilation at the horizon? I’d have to do some research before answering. Until then please disregard the second part of my comment above. However the first part still applies. Yes, in our view there is nothing inside a black hole, but if any mass were there, it would produce no gravity outside anyway. Nothing inside can affect anything outside simply because the inside and outside are separated by time, not just space. Everything outside is in the past relative to anything inside. – safesphere Aug 10 '23 at 05:06
  • @D.Halsey The Schwarzschild singularity is an infinitely long spacelike line, meaning it is a moment of time that happens everywhere in space, like a midnight with no tomorrow. Does midnight have angular momentum or mass? No. Weird, but simple (see the chart in my link above). The Kerr singularity is a timelike ring, meaning that it actually is an object in space that exists forever in time and you can pass it by without ending up there. However the time this singularity exists in has no relation to our time outside, so any properties this singularity may have cannot be manifested outside. – safesphere Aug 10 '23 at 05:19