# How do I feed my black hole?

Strange title I know, but you'll understand why quickly.

I wanted to use a black hole to create absurd amounts of power for hopefully obvious reasons (kind of useful). The problem with natural large black holes is that they tend to be heavy and produce less energy per second than small black holes (or that is what I understand, I would expect the larger surface area to mean it generates more energy but have less surface area to volume to release that energy but that doesn't seem to be the case).

The problem with small black holes is that they disappear on you quickly due to evaporation while you get the energy. So you need to feed them. A very small black hole evaporates almost instantly in a nuclear explosion and they are tiny with Schwarzschild radii measured at the Planck scale.

In my story they use a chamber that suspends the black hole, can absorb all the energy of the black hole and convert it into power to run things. And the chamber has another function, it can activate and at no energy cost can stop the black hole from evaporating in its entirety. There is no grey area here, it can only stop evaporation completely or normal evaporation occurs. This is all important for the story as part of it takes place inside the room of the black hole and the plot requires at least these things!

The question I have: what is the reasonable maximum power output I can get while feeding the black hole using this system while turning the black hole off as little as possible (so you can limit the amount of black holes you need to bring along which can affect the story).

The answer has to explain the following:

• How do you feed the black hole despite it expelling large amounts of energy? (Or why cant you feed it while it is active and need to turn it off to feed it)

• If the black hole needs to be deactivated to be fed, how long does it need to be offline to properly feed it?

• What energy output is this black hole going to give, and based on that how much material needs to be thrown in per second to keep the black hole at that size?

• How do you feed the black hole based on its size?

• What material would be best suited to feed the black hole? (I assume it'll be heavy elements in cases where the black hole is mere hundreds of kilo's in weight)

The best answer will have the most energy output with the least time offline (not amount of times offline, amount of total time offline). In case it is important, the method that offers the highest energy output on average over one cycle of activation, deactivation and feeding is the best answer. For clarity: highest means the most energy per second, rather than the most energy across the entire cycle.

• – Logan R. Kearsley Aug 25 at 21:58
• @LoganR.Kearsley thanks, although they didn't have the answer as well it is very enlightening (as far as such math and subjects go). I don't expect it but maaaaybe someone has the answer to feeding sub-molecular Schwartzchild radius Black Holes now...? – Demigan Aug 26 at 6:55
• The way I see it, a miniature black hole could probably only be created by a civilization already working with Dyson-sphere levels of energy, as that's about what you'd need to create one. As such, using a black hole as a power source is less about GENERATING power so much as STORING it. You've built a Dyson sphere/swarm already, now you need something to do with all that excess energy, so you mass-produce black holes and use them to power sublight generational ships to other stars or something else that requires a ridiculous amount of portable energy. – FlyingLemmingSoup Aug 26 at 6:57
• I think Schwarzschild's name is the single most often misspelled one. – ths Aug 26 at 9:08
• The plural of "radius" is "radii", not "radius's" (or even "radiuses" for that matter). It's too small of an edit for me to make. – Monty Harder Aug 26 at 15:51

In practice, you may not need to feed it at all - as long as you can deal with a sufficiently hefty mini black-hole. According to wiki a black hole with mass $$M$$ measured in kg will have:

Power output $$P = 3.56\times 10^{32} M^{-2}$$ Watts.

Evaporation time $$T = 2.67\times 10^{-24} M^3$$ years.

So a tidy little black hole of mass $$10^{11}$$ kg will provide you with a continuous 36 GigaWatts. Its evaporation time is 2.7 billion years and it will eventually ramp up its power output, but for the first 2000 years or so the power output won't change significantly. This means that the simplest option is to just use it continuously (and leave it for your great-great-$$great^{120000000}$$ grand children to find a safe way to safely dispose of the remnant.

Of course you could feed it approximately 400 micrograms of matter for every second you leave it 'switched on', in which case it will effectively last forever. But the problem is that your black-hole has a diameter of approximately $$1.5\times 10^{-16}$$ m, which is about 10 times smaller than a proton, so the absorption cross-section for any kind of matter will be very small. So it is not obvious that you could actually 'feed' it in any practical way - even if your chamber could hand-wavingly stop it from evaporating during the feeding process.

As well as size of the Schwartzchild radius being problematical, the radiation pressure at its surface (about $$5 \times10^{63}$$ W/m^2) would definitely be a hinderance to poking matter into it, if you didn't stop it from evaporating during feeding.

• I think it's more a problem with hitting it with 40 micrograms of matter every second. You would need something like an extremely powerful coherent neutron beam, extremely precisely aimed, and not disrupted by the 3.6 gigawatts of radiation (in a variety of forms) pouring off the black hole. Resorting to an electric charge differential (say, a very negatively charged black hole bombarded by a proton beam) is probably no better an option, since a black hole with enough charge to help is probably not physically possible, since it would have a naked singularity. – SudoSedWinifred Aug 26 at 8:16
• @SudoSedWinifred Firing a proton beam simply is imparting kinetic energy onto a bunch of protons. Looking at the system as a whole, the amount of mass is actually decreasing due to Hawking radiation. Because of mass-energy equivalence, taking the radiation into account as part of the system would leave the amount of mass constant; this system transforms matter into energy; it doesn't extract kinetic energy from protons. – rytan451 Aug 26 at 8:29
• @Penguino Something seems off about the Power formula. Maybe it should be $M^{-2}$ instead of $M^2$? Otherwise you get way more power. Even then I am not sure how you get 3.6 GigaWatt. Maybe you meant TeraWatt? – Marcel Krüger Aug 26 at 9:22
• I think the issue is that you're converting kilograms to grams when plugging it into the formulas, but the formulas are already written for kilograms. For a time to evaporation of 2700 years, I'm calculating (2.66532*10^-24) * x^3 = 2700, or x = 1.00432 * 10^9 kg, which is almost exactly 3 orders of magnitude off the mass you used. – Josh Eller Aug 26 at 16:35
• Furthermore, using that mass of 10^9 kg and the formulas on the wiki page, I'm getting an initial power output of 356 terrawatts! According to WolframAlpha, that's roughly 150 times the current average global power consumption. – Josh Eller Aug 26 at 16:43