# How might a civilization artificially keep a supermassive black hole electrically charged for the lifetime of the black hole?

A civilization creates a supermassive black hole with an electric charge such that the radius of its inner horizon is 3/4 the radius of its outer horizon. In nature, a black hole with this much electric charge would be neutralized by particles of opposite electric charge falling into the black hole. How might this civilization artificially prevent such particles from getting into the black hole in order to keep the black hole significantly electrically charged for its entire lifetime?

• Why do you pose any problem at all?! These guys are demigods who can create SM black holes. The least of their problems is to keep them stable. Just use their magic and handwavium to make it work – Valerio Pastore Jun 23 '18 at 6:09
• @valerio creating the hole just requires dragging enough mass in and letting it clump together. Our current tech could do that, given enough time. Controlling the charge is very different tech. – SRM Jun 23 '18 at 6:38
• Hundred of thousands of BILLIONS the mass of the sun. This is a SMBH. It is the core of galaxies. You don't just throw stuff in a single point until you get the desired dessert. en.wikipedia.org/wiki/Supermassive_black_hole – Valerio Pastore Jun 23 '18 at 6:42

## 1 Answer

How might this civilization artificially prevent such particles from getting into the black hole in order to keep the black hole significantly electrically charged for its entire lifetime?

They don't bother trying this.

They simply add charge to compensate for the mass increase.

Trying to prevent charge entering would require a lot more effort than adding charged material. At least it would for a civilization that can build a supermassive charged black hole in the first place as they must be have been able to generate insane quantities of charged material "easily".

a supermassive black hole with an electric charge such that the radius of its inner horizon is 3/4 the radius of its outer horizon

Using the Reissner-Nordström metric for a charged and non-rotating black hole we have for the two radii ;

$$r_{out} = \frac 1 2 \left( r_s + \sqrt{r_s^2-4r_Q^2} \right)$$

$$r_{in} = \frac 1 2 \left( r_s - \sqrt{r_s^2-4r_Q^2} \right)$$

As you require $r_{in} = \frac 3 4 r_{out}$

$$4r_s - 4\sqrt{r_s^2-4r_Q^2} = 3r_s + 3\sqrt{r_s^2-4r_Q^2}$$

$$r_s = 7\sqrt{r_s^2-4r_Q^2}$$

$$r_s^2 = 49 r_s^2 - (49)(4) r_Q^2$$

$$48r_s^2 = 196r_Q^2$$

$$12r_s^2 = 49r_Q^2$$

$$r_Q = \frac {\sqrt{12}}{7} \approx 0.4949 \,r_s \approx \frac 1 2 r_s$$

Now :

$$r_s := \frac {2GM}{c^2}$$

and

$$r_Q^2 := \frac {Q^2G}{4\pi \epsilon_0 c^4}$$

So your black hole builders simply need to maintain a charge given by :

$$Q = \left( \sqrt{\frac{192}{49} \pi \epsilon_0 G} \right) M$$

So they can accomplish their goal by simply adding charge as the black hole mass increases to match this equation.

The constant $\sqrt{\frac{192}{49} \pi \epsilon_0 G} \approx 8.53 \times 10^{-11}$ Coulombs per kilogram, so the amount of charge to add is pretty small per unit mass.

As they detect the charge and mass easily from the effects of the gravitational and EM fields, it's pretty simple to track and keep the mass and charge in balance.

EDIT :

Note that we know of no charged massless particles so to add charge we are also adding mass.

Let's say $a := \sqrt{\frac{192}{49} \pi \epsilon_0 G}$.

So we start with a black hole where this is true :

$$Q_0 = aM_0$$

and we add a mass $M_x$ and $N$ electrons of mass $m_e$ each and charge $e$ so we need :

$$Q_0+Ne = aM_0 + aNm_e + aM_x$$

so we need :

$$Ne = Nam_e + aM_x$$

or

$$N = \frac {aM_x}{e-am_e} \approx \frac {aM_x} e$$

What that means is that for every kilo of uncharged mass that the black hole absorbs your advanced aliens will need to add about $5 \times 10^8$ electrons. A kilo of ice contains about $5.7 \times 10^{26}$ electrons, so this is unlikely to pose a problem for black hole engineers.