Double Hawking temperature in de Sitter Universe and cosmological constant problem

Understand

As distinct from the black hole physics, the de Sitter thermodynamics is not determined by the cosmological horizon, the effective temperature differs from the Hawking temperature.

• In particular, the atom in the de Sitter universe experiences thermal activation corresponding to the local temperature, which is twice larger than the Hawking temperature, $T_{\rm loc}=2T_{\rm Hawking}$.
• The same double Hawking temperature describes the decay of massive scalar field in the de Sitter universe.
• The reason, why the local temperature is exactly twice the Hawking temperature, follows from the geometry of the de Sitter spacetime.