# How can we detect a small black hole?

We’re in the late 2020s and we’ve all the planned future telescopes - WFIRST, JWST, Giant Magellan Telescope etc - and technologies.

There’s a small black hole in the outer solar system. It has a radius of about 2.5 attometers and a mass of about 1.5 millions of tonnes.

(How) can we detect such a small black hole with late 2020s technology bearing in mind that

• Someone told us its exact location: L4/L5 Sun–Jupiter Lagrangian point;
• Someone over there could draw attention on it (I don’t know, aher, feeding it for example?);
• We'll send some cool stuff over there;
• The lifespan of the BH is pretty long.
• You made it unphysical with the last bullet... – L.Dutch Jul 24 '17 at 16:37
• If the lifespan of the object in question is "pretty long" then it is not a black hole. Since it's not a black hole, it won't behave like a black hole. How can anybody say how we could detect an exotic object with undescribed properties? – AlexP Jul 24 '17 at 16:38
• Paper of the last bullet: arxiv.org/pdf/0908.1803.pdf (table at page 15) – Lupetto Jul 24 '17 at 16:39
• To clarify the "lifespan" controversy - the given mass (1.5 E12 kg) is sufficient to qualify as a "primordial" black hole, i.e. its lifespan would be large than the age of the universe. So, this black hole is not going to go out with a bang anytime soon. – Alexander Jul 24 '17 at 17:30
• @Alexander You're off by a factor of 1000 for the mass (you've stated 1.5 billion tonnes), it should be ~1.5e9 kg which agrees with the radius of 2.5 attometers. Still, the evaporation time of a black hole that size is around 12700 years. So it'd stick around for a while. – Kyle Jul 24 '17 at 17:37

# Look for the gamma rays

Let's start off with a couple equations:

The Schwarzschild radius of a black hole:

$r_s = {2 G M\over c^2}$

$P = \frac{\hbar c^6}{15360\pi G^2 M^2} = \frac{\hbar c^2}{3840\pi r_s^2}$

And we'll make use of the evaporation time too:

$t_{ev} = \frac{5120\pi G^2 M^3}{\hbar c^4} = \frac{640 \pi c^2 r_s^3}{\hbar G}$

So if we plug in 2.5 attometer for $r_s$, we can find the power emitted by your black hole. It comes out to be around 1.257 * 10^14 Watts or 125.7 TW. Also plugging into the evaporation time we can find that this black hole will have a lifetime of around 12700 years, so there's plenty of time to spot this thing (especially since it'll be continually increasing in brightness unless something/someone is feeding it).

This ends up being equivalent to a black-body with a temperature of around 7.3 * 10^13 K. So this thing is predominantly radiating gamma radiation.

But how bright is this really? For that we need to work out its apparent magnitude. We can do this by computing its intensity compared to the sun. I'll assume the black hole is about 10 billion km away (for reference Pluto is about 7.5 billion km from Earth on average). This is a bit of an abuse of apparent magnitude since it usually refers to visible light and I'm using the entire radiated power of the black hole.

Using that as a reference, I get an apparent magnitude of about 13 (the sun's apparent magnitude is -27 and this black hole is about 40 steps above that, so 13).

That makes this thing about 2.5 times brighter than Pluto or about as bright as this quasar, so you definitely won't be seeing it with the naked eye. However, it should be pretty noticeable if you point a gamma ray detector at it I would think.

• Yeah, the IONDS would pick up on this showing up for sure – Shalvenay Jul 24 '17 at 22:54
• Physics is great. It's called "Black hole" and you compare it with something called a "black-body", yet it can be detected because it is so bright. – Fabian Röling Jul 25 '17 at 10:14
• Uhm, gamma rays... sounds dangerous, isn't it? – Lupetto Jul 25 '17 at 18:14
• @Lupetto Well, it's a black hole, nothing about them is safe. – Kyle Jul 25 '17 at 18:33
• The black hole is in Jupiter's L4 or L5 point, so it's only 778 $\pm$ 1.5 million km from Earth. That's about 12.85 times closer, or 165 times brighter. If I did my math right, that's about 5.5 magnitude lower, or 7.45 absolute magnitude. According to wikipedia, that's right at the limit of visible magnitude (though obviously, this is gamma rays we can't see anyways). – MichaelS Jul 26 '17 at 23:30

Such a small black hole would evaporate pretty quickly emitting a flash of Hawking radiation. If you are quick, you can catch it.

• You are right. I added a fourth point on the list. – Lupetto Jul 24 '17 at 16:35

With a black hole that small and that close, you'd easily be able to see it as an gamma ray source. According to your sheet, the thing is outputting 20 Pentawatts of hawking radiation. Which is more 4000x more power than the total power the earth gets from from the sun.

### Update: don't do this. It won't work.

Use a black and decker laser range finder. jk

But since it is so close, I would shine a laser toward the blackhole. Get it to bend around the blackhole and come back to us. You're not going to get it perfect but you should be able to detect something of a signal. What I like about this way is that it's active detection and not passive detection (looking for light already out there to bend).

• The black hole is tiny, the amount of light that gets bent that much will be far less than the amount reflected if a single speck of dust was in the way. – Donald Hobson Jul 24 '17 at 19:29
• That's very true. I figure since they know where it is they could probably direct a reasonably strong pulse towards it, maybe even haphazardly. Voyager 1 has a 22 watt transmitter, after all the dust and beam divergence I can't imagine that's too strong either, but we can detect it with large dishes. I don't know the math. – fet Jul 25 '17 at 0:23
• Still won't work. The bending is related to the mass (and the distance from the black hole) - Pluto is much (ten billion times) more massive than this thing. – Martin Bonner Jul 25 '17 at 10:58
• @MartinBonner, interesting. Would you mind explaining? I don't quite understand. :) – fet Jul 25 '17 at 14:12
• There nothing particularly magical about a black hole, it's just a mass with a very strong gravitational field because it is comparatively small. The point is, that things like bending light only depends on the field - and this falls off by 1/r². At a distance of 1 metre from the centre of the BH, the BH only exerts a pull of 1E-11 N on a mass of 1kg. – Martin Bonner Jul 25 '17 at 14:49