The Evaporation of Black Holes
In 1974, Stephen Hawking shocked the physics world by proving that black holes aren't completely black. Due to the bizarre rules of quantum mechanics operating right at the edge of the event horizon, black holes actually emit a very faint glow of thermal radiation, now known as Hawking Radiation.
Because they are emitting energy, they must be losing mass ($E=mc^2$). If a black hole doesn't consume new matter to replace what it radiates away, it will eventually shrink and "evaporate" entirely in a massive flash of gamma rays.
The Strange Rules of Temperature
Hawking's equation reveals an inverse relationship: the heavier the black hole, the colder it is.
- A supermassive black hole at the center of a galaxy is incredibly cold (billionths of a degree above absolute zero).
- A tiny, microscopic black hole would be blindingly hot, burning up almost instantly.
The Formula
Example Calculation
Calculate the temperature of a "stellar-mass" black hole that is exactly $5$ times heavier than our sun ($Mass \approx 10^{31} , \text{kg}$).
(Note: The constants $\hbar, c, G, k_B$ combine to form a constant multiplier. For quick math, $T \approx 1.227 \times 10^{23} / Mass$)
- Divide Constant by Mass: $1.227 \times 10^{23} / 10^{31} = 1.227 \times 10^{-8} , \text{Kelvin}$.
The temperature of a standard black hole is a tiny fraction of one billionth of a degree above absolute zero.