I would be very curious if Kerr black holes emit Hawking radiation at the same temperature in the equatorial bulges and in their polar regions. I've been looking some reference for this for a couple of months now but i haven't been successful yet.

According to this answer, Too great Hawking radiation could thwart mass from being feed into a black hole, so this begs the question: Could possibly a (very small) Kerr black hole be able to accept infalling matter if it falls along the rotational axis? Possibly, since 4D Kerr black holes angular momentum are bounded by extremality condition, the possible difference in temperature between polar and equatorial regions will not be wide enough to be interesting (possibly not bigger than an order of magnitude), but it would still be interesting to know how much is the temperature ratio. It could very well be zero, or more intriguingly, could depend on the black hole mass. But in any case, i doubt that the ergosphere region will not interact nontrivially to boost equatorial Hawking radiation to some degree


  • 2
    $\begingroup$ I think the most relevant question is this: physics.stackexchange.com/q/44323 In that question, the formula for total Hawking radiation luminosity for a spinning, charged, black hole was obtained. In obtaining this, I'm sure that ability to find the equatorial and polar luminosities was requisite. This information would then be critical to your current question. $\endgroup$ Commented Dec 14, 2012 at 19:35
  • $\begingroup$ Indeed, as it approaches extremality, the black hole cools down overall. But i haven't found an expression for the angular distribution of the radiation for these black holes. I would think too it should be a requisite to know the overall radiation rate $\endgroup$
    – lurscher
    Commented Dec 14, 2012 at 19:52
  • 1
    $\begingroup$ A non-quantum paper that crops up in my searches is here (alas, also paywalled). It calculates radiation of classical waves out of the black hole, as seen by an observer at infinity. Teukolsky notes that he gives partial solutions for the near-field, which got me wondering: Do we necessarily have to know where on the event horizon a particle is emitted in order to know the total rate? Global considerations might get you differential cross sections for observers far enough away that the BH is a point source. $\endgroup$
    – user10851
    Commented Jan 24, 2013 at 1:13

1 Answer 1


I don't have a full answer to your question, but your question did make me realize that the Hawking radiation must be carrying away angular momentum from a rotating Kerr black hole. If it only carried away energy (mass) the Kerr black hole would become extremal as it lost mass but kept the same angular momentum. This, I think, proves that there has to be asymmetries in the Hawking radiation from a Kerr black hole.

One way to carry away angular momentum would be for a higher flux or energy of Hawking radiation in the forward direction of rotation at the equator. I think this implies that limb coming towards you would look like it is at a higher temperature that the limb that is receding from your point of view. Another way to carry away angular momentum would be to emit partially (circularly) polarized radiation from the polar regions but then these would not look like black body radiation. I don't know how to calculate the effects, but I would not be surprised if there was a temperature variation on the event horizon. But I could be wrong.

"lurscher" gave a link to an article that answers this question. The abstract says:

Particle emission rates from a black hole. II. Massless particles from a rotating hole

Don N. Page W. K. Kellogg Radiation Laboratory, California Institute of Technology, Pasadena, California 91125

The calculations of the first paper of this series (for nonrotating black holes) are extended to the emission rates of massless or nearly massless particles from a rotating hole and the consequent evolution of the hole. The power emitted increases as a function of the angular momentum of the hole, for a given mass, by factors of up to 13.35 for neutrinos, 107.5 for photons, and 26 380 for gravitons. Angular momentum is emitted several times faster than energy, so a rapidly rotating black hole spins down to a nearly nonrotating state before most of its mass has been given up. The third law of black-hole mechanics is proved for small perturbations of an uncharged hole, showing that it is impossible to spin up a hole to the extreme Kerr configuration. If a hole is rotating fast enough, its area and entropy initially increase with time (at an infinite rate for the extreme Kerr configuration) as heat flows into the hole from particle pairs created in the ergosphere. As the rotation decreases, the thermal emission becomes dominant, drawing heat out of the hole and decreasing its area. The lifetime of a black hole of a given mass varies with the initial rotation by a factor of only 2.0 to 2.7 (depending upon which particle species are emitted). If a nonrotating primordial black hole with initial mass 5 × 1014 g would have just decayed away within the present age of the universe, a hole created maximally rotating would have just died if its initial mass were about 7 × 1014 g. Primordial black holes created with larger masses would still exist today, but they would have a maximum rotation rate determined uniquely by the present mass. If they are small enough today to be emitting many hadrons, they are predicted to be very nearly nonrotating.

  • $\begingroup$ I don't know either, but this reference claims "We show that the 4-dimensional Kerr-Newman metric, which has a spherically nonsymmetric geometry, becomes an effectively 2-dimensional spherically symmetric metric by using the technique of the dimensional reduction near the horizon)" $\endgroup$
    – twistor59
    Commented Dec 14, 2012 at 20:21
  • $\begingroup$ @twistor59 and FrankH, it seems that this article has an answer for this question: prd.aps.org/abstract/PRD/v14/i12/p3260_1, but i don't have access to it $\endgroup$
    – lurscher
    Commented Jan 23, 2013 at 17:53
  • $\begingroup$ @lurscher That linked article (part 2 in the series) unfortunately just gives an integrated emission as a function of $M$, $a$, and the quantum numbers of whatever particle you are considering being emitted, without derivation. It cites part 1 of the series, but that in turn declares the integrated rate to be the result of "Hawking's calculation." $\endgroup$
    – user10851
    Commented Jan 24, 2013 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.