Hawking Temperature
Temperature of the radiation a black hole emits.
Formula
Variables
Example
A solar-mass black hole is a mere 60 billionths of a kelvin
How it works
This formula gives the effective temperature of Hawking radiation emitted by a non-rotating, uncharged black hole. It comes from combining quantum-field effects near the event horizon with gravity: virtual particle effects and curved spacetime make the black hole behave thermally, with smaller black holes radiating more strongly because their horizons are hotter. The inverse dependence on mass means the temperature rises as the black hole gets lighter.
When to use it
Use this when estimating the thermal spectrum or evaporation rate of an ideal Schwarzschild black hole. It is also useful in theoretical astrophysics and quantum gravity problems where you need the black hole's characteristic temperature from its mass.
Worked example, step by step
Find the Hawking temperature of a black hole with mass 10^12 kg. T = ħ c^3 / (8π G M k_B) Using ħ = 1.055×10^-34 J·s, c = 2.998×10^8 m/s, G = 6.674×10^-11 N·m^2/kg^2, k_B = 1.381×10^-23 J/K, M = 1.0×10^12 kg First compute the constant numerator factor: ħ c^3 ≈ (1.055×10^-34)(2.998×10^8)^3 ≈ 2.84×10^-9 Compute the denominator constant: 8π G k_B ≈ 8π(6.674×10^-11)(1.381×10^-23) ≈ 2.32×10^-32 Now divide by the mass: T ≈ (2.84×10^-9) / (2.32×10^-32 × 1.0×10^12) T ≈ 1.22×10^11 K Final answer: the Hawking temperature is about 1.2×10^11 kelvin.
Common mistakes to avoid
- Using the black hole's radius directly instead of its mass in this formula
- Forgetting that the temperature is inversely proportional to mass, so smaller black holes are hotter
- Applying the formula to spinning or charged black holes without modification
Did You Know?
Smaller black holes are hotter, so they evaporate ever faster — the final instant is a burst of radiation.
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