# Blandford-Znajek process on micro black holes

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I'm interested in studying the Blandford-Znajek process and how the rate of power generation changes when the spinning black hole mass is small ($M_b < 10^{16}$ kilograms). It is known that black holes tend to become harder to feed via accretion as their masses become smaller, but my understanding of the BZ process is that is not directly reliant on accretion, but on the shape of the surrounding toroidal magnetic field. Of course, is not clear how such field could be created other than by an accreting disk

Any good, self-containing references are highly appreciated as always

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(This felt like a too long comment so I post it as an answer)

The basic equations are usually expressed in geometrized units so it's no problem to rescale the mass and get the respective quantities (this can really be obtained even from the original article). Then you have the required magnetic field $\sim L^{1/2}$ with the output luminosity and $\sim M^{-1}$ with the black-hole mass.

For supermassive black holes you usually get $\sim 0.01 T$ magnetic fields but for the tiny black hole with mass $\sim 10^{-23}$ times that of the supermassive to even flicker, I fear the magnetic fields would have to be monstrous. Additionally, I do not really see what sense does it make to study the effect in concrete situations without having a handle on the external supporting current (aka the accretion disc).

answered Jun 9, 2015 by (1,645 points)
edited Jun 9, 2015 by Void

The term "micro black hole" confuses me a bit, as usually it means a quantum mechanical black hole. Is the effect even defined for such objects?

@Dilaton Sorry, I didn't know it was a thing. The usual nomenclature is stellar-mass, supermassive and intermediate. I just wanted to underline that $10^{16} kg$ is simply "micro" when compared to the supermassive or even "smallest astrophysical", stellar.

"I fear the magnetic fields would have to be monstrous"
why? If you see ineq. 2.8 for the minimum magnetic field, $B >\frac{M^{1/4} M_s^{1/2} }{ a^{3/4} }$, so the magnetic field *falls* as the quartic root of the mass

Yes, the vacuum-breaking $B$ does behave this way. But the energy extraction (aka $L$) is computed independently. What I am referring to is the rather brute approximation as discussed towards the end of Section 8, unfortunately no equation numbering, the one before last equation. Obviously, it depends on how you define a sufficient luminosity for the BH to "even flicker". I was thinking something like a linear mass-luminosity rescaling, which would require $\sim 10^8 T$ magnetic fields. If you just stuck with $B \sim 0.01T$ the output luminosity would be $\sim 10^{-46}$ that of an active galactic nucleus. Which could be enough if you are considering Blanford-Znajek to power some local effects but observationally this would probably be irrelevant.

the way these guys write some expressions with a minus-one exponent obfuscates unnecesarily expressions. In fact, that expression for $B$ seems to be independent of $M$, as it appears once with a positive one exponent, and once with a negative one exponent. If I group all dependences, I get $B = 0.2 \times L_H^{1/2} \times 10^{-38} W \times 10^{9} \frac{ M_s }{a}$ T

Haha, then that is a completely different question :) What you then need is just a sufficient supporting current to break the vacuum and you have $\dot{M} \sim L$ under control. The only thing you need is to achieve sufficient accretion efficiency but you can certainly get above $0.4\%$, which is the case of fusion - even a pessimistic guess would be units of percent, tens of percent if you are optimistic (but don't ask me how to power a spaceship with disperse radiation around a black hole).
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