Hawking radiation and rotating Black holes

[Hiranya Pasan]

Hawking radiation formula shows the fact that when charge and angular momentum increases in a Kerr-Newman black hole (angular momentum in Kerr black hole) Hawking radiation decreases.

Can someone explain this?
Thank you.

 

[PeterDonis]

Hawking radiation formula shows the fact that when charge and angular momentum increases in a Kerr-Newman black hole (angular momentum in Kerr black hole) Hawking radiation decreases.

Do you have a reference for this statement?

 

[Hiranya Pasan]

Do you have a reference for this statement?

https://physics.stackexchange.com/q...e-lose-mass-via-hawking-radiation/44327#44327

 

[PeterDonis]

Stack Exchange is not necessarily a reliable source. Do you have a textbook or peer-reviewed paper that derives this result?

 

[PeterDonis]

when charge and angular momentum increases

You also need to be more precise about what you mean by this. By what process are you assuming that the charge or angular momentum increases? What effect does this process have on other black hole parameters, such as the mass?

 

[stevebd1]

A bit late to this one but thought it was worth making a contribution. Hawking radiation is related to the Killing surface gravity and the Killing surface gravity is predicted to reduce the more charge and spin are increased with it approaching zero as a black hole approaches maximal ((a²+q²)/M²=1). Below are a couple of sources but there are more out there. First the equation for Hawking radiation in geometric units-
$$T_{\text{H}}=\frac{\kappa}{2\pi}$$
where $\kappa$ is the surface gravity

One source defines $\kappa$ as
$$\kappa=4\pi\frac{\sqrt{M^2-q^2-a^2}}{A}$$
where $A=4\pi\left(2M^2-q^2+2M\sqrt{M^2-q^2-a^2}\right)$, $A$ being the surface area of the BH, $M$ is mass, $q$ is charge and $a$ is spin (in the paper $a$ is written as $J/M$ which is equivalent).

source-
Quantum Aspects of Black Holes by Dorothea Deeg
https://edoc.ub.uni-muenchen.de/6024/1/Deeg_Dorothea.pdf equations 1.23 & 1.25

You can see here that as spin and charge increase, the Killing surface gravity reduces and so does the Hawking radiation. Another source (which only deals with spin) defines $\kappa$ as
$$\kappa=4\pi\frac{\mu}{A}$$
where $A=4\pi\left[2M(M+\mu)\right]$ and $\mu=\sqrt{M^2-a^2}$

source-
Hawking radiation screening and Penrose process shielding in the Kerr black hole by Eamon Mc Caughey
https://arxiv.org/abs/1603.08774 equations 33, 34, 35

For want of another source, the Wikipedia entry for the Kerr-Newman solution for surface gravity shows the same thing, that charge and spin result in the surface gravity reducing-
https://en.wikipedia.org/wiki/Surface_gravity#Kerr–Newman_solution

 

[PeterDonis]

You can see here that as spin and charge increase, the Killing surface gravity reduces

Not necessarily. Changing the spin or charge also changes the area. And any process that adds spin or charge to the hole also adds mass. It's not at all clear that the end result of such a process will always be a decrease in the surface gravity of the hole.

 

[sweet springs]

I do not know evaporation of Kerr BH, but if so, after evaporation nothing remains there, emitted particles or lights should carry out all angular momentum BH has.

 

[stevebd1]

Not necessarily. Changing the spin or charge also changes the area. And any process that adds spin or charge to the hole also adds mass. It's not at all clear that the end result of such a process will always be a decrease in the surface gravity of the hole.

The area $(A)$ does also decrease but not as fast as the top part of the equation for $\kappa$. For example using M=4000 m, for a static black hole $\kappa=1/4M$ (which is what the equations in post #6 reduce to when $a=0$) and $\kappa=\text{6.25e-5}$, for a black hole with a spin parameter of a/M=0.99, a=3960 and $\kappa=\text{1.545e-5}$, a reduction.

You say that spin and charge add mass to the black hole but M already takes this into account when you look at the equation for irreducible mass-
$$M^2=\frac{J^2}{4M_{ir}^{2}}+\left(\frac{Q^2}{4M_{ir}}+M_{ir}\right)^2$$
where $J=aM$ and $M_{ir}=1/2\sqrt{\left(M+\sqrt{M^2-Q^2-a^2}\right)^2+a^2}$, the irreducible mass being the mass of the black hole after any spin or charge has been removed, source. If they equations for the Killing surface gravity used $M_{ir}$ then I'd say something was amiss but the use $M$ which includes for the extra mass due to spin and charge.

Here's another source that shows the reduction directly related to Hawking radiation (again only relating to spin), which is a combination of the equations in post #6, giving the same results, again showing reduction-
$$T=\frac{\sqrt{M^2-a^2}}{4\pi M(M+\sqrt{M^2-a^2})}$$
Source-
Hawking Radiation from Rotating Black Holes and Gravitational Anomalies by Keiju Murata, Jiro Soda
https://arxiv.org/abs/hep-th/0606069v2 equation 34

There seems to be some support for the idea that Hawking radiation decreases as spin and charge increase. I think it contributes to the notion that a maximal black hole ((a²+q²)/M²=1) cannot exist, that you cannot have a naked singularity due to the cosmic censorship conjecture and that $\kappa=0$ would violate the third law of black hole thermodynamics, https://www.physicsforums.com/threads/black-hole-thermodynamics.762982/. I'm aware that some metrics can play mathematical tricks, there's debate about how realistic the inner Cauchy horizon is in charged and spinning black holes, white holes are another trick but I think there's something in the reduction of the surface gravity and Hawking radiation due to spin & charge and there seems to be mathematical support for it.

 

[PeterDonis]

You say that spin and charge add mass to the black hole

No, I said that any process which adds spin or charge also adds mass, because whatever thing is carrying the spin or charge is also carrying mass.

M already takes this into account

No, it doesn't, because $M$ is the mass of the hole before the process that adds spin or charge; after that process, the new mass of the hole is $M + m$, where $m$ is the mass of whatever thing carried the spin or charge that was added.

 

[stevebd1]

No, it doesn't, because $M$ is the mass of the hole before the process that adds spin or charge; after that process, the new mass of the hole is $M + m$, where $m$ is the mass of whatever thing carried the spin or charge that was added.

I understand what you're saying here and if I'm honest, I knew this and it wasn't the point I was trying to make. I know that if you're lucky enough to find a black hole with a spin parameter of 0.999 and you think you could just 'push it over the edge' by adding a little more spin or charge, this won't happen because by what ever process you add spin and mass, you will also add to M and you will always fall short of maximal (unless there's a process that can add spin and charge without adding to M which isn't currently known of). The point I was making is that say you have two black holes that both read from a distance as having a mass of 10 sol, one is as near static as possible (for arguments sake) and the other has a spin parameter of a=0.95, then the one with spin will radiate less Hawking radiation than the static black hole-

(Reasonably) static black hole
$$M=14767.7558$$
then $\kappa=1.692877e-5$ and $T_H=2.694298e-6$

Rotating black hole (a=0.95)
$$M=14767.7558$$
$$a=14029.36801$$
then $\kappa=8.056405e-6$ and $T=1.282217e-6$
(the BH would have an irreducible mass $(M_{ir})$ of 11962.10964 implying that approx. 19% of the BHs mass observed from a distance was due to spin)

The point being that if charge and spin are already present in the black hole, then the BH with charge and spin will have a lower temperature than a static black hole with a perceived same mass (which is what I think the OP was getting at). I get your point that decreasing the temperature further by adding more spin and charge is ambiguous due to the fact that you be would adding to M also.

A few more sources, one for Killing surface gravity-
$$\kappa=\pm\frac{\sqrt{M^2-a^2}}{2M(M\pm\sqrt{M^2-a^2})}$$
from the paper-
'..where the plus sign applies to the event horizon ${r = r_+}$, and the minus sign should be used for the Cauchy horizon ${r = r_−}$. In the extreme cases $m = \pm a$ only the plus sign is relevant. We see that $\kappa$ vanishes then, and is not zero otherwise.'

Source-
The Geometry of Black Holes by Piotr T. Chru´sciel
https://homepage.univie.ac.at/piotr.chrusciel/teaching/Black Holes/BlackHolesViennaJanuary2015.pdf equation 1.6.34

Another source temperature due to charge only-
$$T=\frac {1}{2\pi}\frac{\sqrt{M^2-Q^2}}{(M+\sqrt{M^2-Q^2})^2}$$

Source-
Entropy is Conserved in Hawking Radiation as Tunneling: a Revisit of the Black Hole Information Loss Paradox by Baocheng Zhang, Qing-yu Cai, Ming-sheng Zhan, Li You
https://arxiv.org/abs/0906.5033 between equations 20 and 21

 

[PeterDonis]

The point I was making is that say you have two black holes that both read from a distance as having a mass of 10 sol, one is as near static as possible (for arguments sake) and the other has a spin parameter of a=0.95, then the one with spin will radiate less Hawking radiation than the static black hole

Yes, agreed.

the BH would have an irreducible mass ($M_{ir}$) of 11962.10964 implying that approx. 19% of the BHs mass observed from a distance was due to spin

I don't think this is a good way of describing it. Mass and spin are different things. It's just that both of them affect the spacetime geometry.

I do think, though, that a good heuristic for the difference in Hawking radiation is that Hawking radiation depends on the irreducible mass, which basically corresponds to the horizon area.

 

[stevebd1]

I don't think this is a good way of describing it. Mass and spin are different things. It's just that both of them affect the spacetime geometry.

'19% of the BHs stress energy tensor was due to spin' might have been a more accurate statement here. On that subject, I was looking to see if there was a simple equation to calculate the energy for a BH that was due to spin. It's possible to derive something from equations for irreducible mass but this is little more than a 'cut & paste' (for rotation only)-

$$KE_{\text{rotation}}=M-\frac{1}{2}\sqrt{r_+^2+a^2}$$

where $r_+$ is the outer event horizon

Oddly enough, the simple equation

$$KE_{\text{rotation}}=\frac{1}{2}J\omega$$

where $\omega$ on this occasion would be $\Omega_+=a/(r_+^2+a^2)$, the angular velocity at the outer event horizon, provides results that are surprisingly close to the 19% (18.999% actual) produced by the irreducible mass equations. On this occasion, the results are 17.193% but it's still not exact. I'm sure there's an equation out there but I'm not aware of it and it's likely the equations for irreducible mass are derived from/based on it (along with a similar equation for charge).

 

[PeterDonis]

'19% of the BHs stress energy tensor was due to spin' might have been a more accurate statement here.

Not really, since a BH is a vacuum solution, so the stress-energy tensor is zero everywhere.

the energy for a BH that was due to spin

There is no such thing. The energy of the hole is its mass, $M$. A rotating black hole is not like an ordinary spinning object where you can separate its energy into "rest energy" and "spin energy".

 

[PeterDonis]

I'm sure there's an equation out there but I'm not aware of it and it's likely the equations for irreducible mass are derived from/based on it

The irreducible mass is really telling you about the area of the hole's horizon; the name "irreducible mass" comes from the second law of thermodynamics as applied to black holes, which says that the total area of all black hole horizons can never decrease. So one way of putting what I think you are trying to get at is that the area of a rotating hole's horizon is smaller, in relation to its mass, than the area of a non-rotating hole's horizon, so the irreducible mass of a rotating hole is less than its actual mass (whereas the two are equal for a non-rotating hole).

 

[stevebd1]

An interesting coincidence that doesn't necessarily mean anything is that the area for a rotating charged black hole is equivalent to the equation for the area for a static black hole using M_ir instead of of M-

$$A_+=4\pi \left(r_+^2+a^2\right)\equiv 16\pi M_{ir}^2$$

where $A=16\pi M^2$ is the conventional equation for a static black hole which the above reduces to. Though a smaller area would normally mean higher radiation for a static black hole, for a charged rotating black hole, it's less and something else is taking place. While there's no guarantee that the effect is local to the horizon It's not a stretch to suggest that maybe the centripetal force brought on by frame dragging and the possible electrostatic repulsion brought on by charge might contribute to less virtual particles being divided near/at the horizon and therefore less radiation. On a slightly different note, while $\kappa$ and T tend to zero depending on spin and charge, the same doesn't seem to apply with entropy which is expressed as S=A/4. A few papers go some way to address this saying that the entropy is the sum of the outer (A₊) and inner horizon (A_-), A_- being negative and the sum of the two tending to zero as the black hole becomes maximal. Some paper also talk about Hawking 'absorption' at the inner horizon-

"The KN black hole has two horizons: the inner (Cauchy) horizon $r_-=M-\epsilon$ and the outer (event) horizon $r_+=M+\epsilon$, where $\epsilon=\sqrt{M^2−Q^2−a^2}$. In particular, the thermodynamics associated with the outer event horizon of the black hole is related to the fundamental process of Hawking radiation. Similarly, one can prove that the thermodynamics of the inner Cauchy horizon is associated with another quantum process of Hawking “absorption” ... Namely, for an observer rest at the infinity he observes a net flux of Hawking radiation outgoing from the event horizon to the infinity, while for an imaginary observer inhabiting in the intrinsic singular region $(0<r<r_-)$, he will observe a flux of Hawking “absorption” ingoing from the intrinsic singular region to the inner horizon. Because a KN black hole is stationary, its outer horizon is in thermal equilibrium with the thermal radiation outside the black hole. We think that its inner horizon is certainly in thermal equilibrium with the thermal radiation in the intrinsic singular region also."

Source-
New formulation of the first law of black hole thermodynamics: a stringy analogy by Shuang-Qing Wu
https://www.sciencedirect.com/science/article/pii/S0370269305000377 paragraph between equations 2 and 3

Another paper that talks about Hawking absorption and combining the outer and inner horizon areas when calculating entropy-
Entropy of Kerr-Newman Black Hole Continuously Goes to Zero when the Hole Changes from Non-extreme Case to Extreme Case by Zhao Zheng, Jian-Yang Zhu and Liu Wen-biao
https://www.researchgate.net/publication/230923684_Entropy_of_Kerr-Newman_Black_Hole_Continuously_Goes_to_Zero_when_the_Hole_Changes_from_Nonextreme_Case_to_Extreme_Case

Two other papers that simply look at combining the two horizons-

Black Holes, Entropy and the Third Law by A. J. Meyer, II
http://arxiv.org/abs/physics/0608080 equation 41

A Proposed Absolute Entropy of Near Extremal Kerr-Newman Black Hole by Hai Lin
http://arxiv.org/abs/gr-qc/0104098 in the abstract and equation 48