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  Blandford-Znajek process on micro black holes

+ 3 like - 0 dislike

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

asked Jun 8, 2015 in Theoretical Physics by CharlesJQuarra (555 points) [ revision history ]

1 Answer

+ 4 like - 0 dislike

(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 Void (1,645 points) [ revision history ]
edited Jun 9, 2015 by Void
Most voted comments show all comments

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.

@CharlesJQuara Yes, the magnetic field would be small but so would the luminosity $L_H \sim \dot{M}$. But beware that a part of this is derived only perturbatively for small spins. It all really depends on what is the motivation of your study. Do you want to create observation models for tiny black holes? Understand their dynamics in some cosmological context?

The motivation is determining if this process could be exploited efficiently by very advanced intelligences in order to create relativistic starships that outperform fusion rockets. There is some people that have considered Hawking evaporation for this purpose, but it has a lot of problems (how to push the black hole with the ship, how to feed the black hole faster than it evaporates, etc.) and I was hoping to examine if those problems persist with the Blandford-Znajek process

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).

Big problems are dragging the still rather massive black hole with you and the fact that the tiny black hole will have very powerful slap forces. Not only would it be rather strenuous on the materials in the quasi-static case but if the distance between the ship and the black hole somehow oscillated during transport, it would slam things around quite a bit. I would recommend a black hole as a futuristic power plant rather than an engine...

Most recent comments show all comments

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

@CharlesJQuarra There is a good reason for that, spin is actually given in units of $M$. That is, e.g. an extremal black hole has spin $a=M$. Being optimistic, you can say that your black hole is near-extremal and $a/M$ is a dimensionless factor $\sim 1$.

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