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What is the equation which gives us the mass loss per unit of time, of a black hole (non-rotating) of a given mass? What does a graph of a range of [BH masses/(mass loss/t)^-1] (?) look like? Is there a startling relation such as S~A/4 (entropy of a BH is proportional to (or equal to, in Planck units) a quarter of the BH area), which inspired the holographic conjecture?

george lastrapes
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It’s very simple: $dM/dt\propto-1/M^2$. In words, the mass loss per unit time is proportional to the inverse square of the mass.

The more mass it loses, the faster it loses it! By the time a black hole shrinks to 10% of its original mass, it is losing mass 100 times faster than it originally was. At 1%, 10,000 times faster. This is why you get an explosive burst at the end.

I can’t understand what graph you are asking for, but you can figure it out yourself from this equation.

Hawking’s discovery of radiation from black holes was startling. This equation is simply a logical consequence of that.

G. Smith
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  • Thank you! As close as I got was, proportional to the square of the curvature of the EH. Might be the same thing-- I have often visualized curved space squeezing out virtual pairs, for some reason. I apologize for the tiresomely simple question, especially to Mr. Nontriviality, but as I told him, I studied the classics rather than physics. – george lastrapes May 14 '19 at 19:53
  • And didn't Beckenstein do most of the heavy lifting for Hawking? – george lastrapes May 14 '19 at 20:03
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    My understanding is that Bekenstein had long been studying the thermodynamics of black holes, but he didn’t yet have a “mechanism” for them to have a temperature. Hawking’s calculation provided that, and then everybody realized that Bekenstein had been on the right track all along. I think his contributions got overshadowed to some degree because of Hawking’s personal story. The work of both men is extremely important to understanding black holes. – G. Smith May 14 '19 at 21:23