# Does Kerr metric have Hawking temperature?

Does Kerr metric with zero charge have Hawking temperature? What is it given by?

I am reading a paper about and it says that the Einstein-Maxwell-Dilaton-Axion black hole and it says that temperature of this black hole is given by $$T =\frac{ r_{+}^2−a^2}{ 4r_{+}(r^2_{+}+2br_{+}+a^2)}$$

So does that mean that the Hawking temperature for Kerr metric is given by $$T =\frac{ r_{+}^2−a^2}{ 4r_{+}(r^2_{+}+a^2)}$$?

• What research have you done on this? What is the definition of "Hawking temperature" here? Do you know how it is arrived at in the Schwarzschild case, and if yes, where does trying to apply the same logic to the Kerr metric fail? – ACuriousMind Dec 17 '16 at 18:57
• $\uparrow$ Which paper? – Qmechanic Dec 17 '16 at 19:30
• @Qmechanic, arxiv.org/abs/1311.4251 – MrDi Dec 17 '16 at 19:33