Hawking Temperature

Hawking Temperature

In 1974, Stephen Hawking showed that black holes are not entirely black — they emit thermal radiation due to quantum effects near the event horizon. This Hawking radiation has a characteristic temperature:

$T_H = \frac{\hbar c^3}{8\pi G M k_B}$

where:

• $\hbar = 1.0546 \times 10^{-34}$ J·s (reduced Planck constant)
• $c = 2.998 \times 10^8$ m/s (speed of light)
• $G = 6.674 \times 10^{-11}$ N·m²/kg² (gravitational constant)
• $M$ = black hole mass in kg
• $k_B = 1.381 \times 10^{-23}$ J/K (Boltzmann constant)

For a solar-mass black hole, $T_H \approx 6 \times 10^{-8}$ K — far colder than the cosmic microwave background (2.7 K). Tiny black holes, however, are extremely hot and evaporate rapidly.

Peak Wavelength of Hawking Radiation

The peak wavelength of the emitted radiation (Wien's law) simplifies to:

$\lambda_{\text{peak}} = \frac{hc}{k_B T_H} = \frac{2\pi\hbar c}{k_B T_H}$

Substituting $T_H$ and simplifying:

$\lambda_{\text{peak}} = \frac{16\pi^2 G M}{c^2}$

Notice this equals $8\pi$ times the Schwarzschild radius — the radiation wavelength is comparable to the black hole size.

Hawking Luminosity

Treating the black hole as a black body radiating from area $A = 4\pi r_s^2$ (where $r_s = 2GM/c^2$), the Stefan–Boltzmann law gives:

$L = \sigma A T_H^4$

with $\sigma = 5.67 \times 10^{-8}$ W/m²/K⁴. Smaller black holes are hotter and far more luminous.

Your Task

Implement three functions. All physical constants must be defined inside each function body.

• hawking_temperature(M) — returns $T_H$ in Kelvin
• hawking_wavelength(M) — returns peak wavelength using the simplified formula $16\pi^2 GM/c^2$
• hawking_luminosity(M) — returns luminosity $L$ using the Stefan–Boltzmann law