# Can anything be seen from the center of the Boötes void? How dark would it be?

Let's say I was at the very center of the enormous Boötes void, way out in deep, deep space. What could I see with the naked eye? I assume I could see no individual stars, but could I resolve any galaxies? If I gazed in the direction of a super-cluster of galaxies would it seem brighter than other directions? How dark would it be compared to, say, the far side of the moon when it is a full moon on earth?

I am told there are, in fact, a few galaxies in the void. So let's say I pick a spot in the void that is as far from any of those galaxies as possible.

• The Bootes void has some galaxies in it, and I don't know if it has much of a centre. IMO this needs to be more specific about the conditions.
– JMac
Apr 17, 2019 at 14:29
• Let's pick the spot that is furthest from any galaxy. Apr 17, 2019 at 14:32
• @PaulYoung so you're saying "if there's a bright center to the Universe, you're at the [point] it's farthest from"? Apr 17, 2019 at 21:47
• @MikeTheLiar seems more like "let's get as far away from as many galaxies as possible" Apr 18, 2019 at 0:38
• As a starter: What would we see from Earth if we removed all those celestial objects that belong to our Galaxy? Apr 19, 2019 at 10:05

### Individual sources

The number density of galaxies in a void is typically an order of magnitude lower than the average in the Universe (e.g. Patiri et al. 2006). In this astronomy.SE post, I estimate the number density of galaxies of magnitude $$M=-17$$ or brighter in the Boötes Void to be $$n \sim 0.004\,\mathrm{Mpc}^{-3}$$, or $$10^{-4}\,\mathrm{Mlyr}^{-3}$$ (i.e. "per cubic mega-light-year"). Hence, the typical distance to a galaxy from a random point in the Boötes Void is $$d = \left( \frac{3}{4\pi n} \right)^{1/3} \simeq 13\,\mathrm{Mlyr}.$$

Although some galaxies will be brighter than $$M=-17$$, the number density declines fast with brightness; for instance, galaxies that are 10 times brighter are roughly 100 times rarer, meaning that they're on average 5 times more distant and hence appear 25 times fainter. On the other hand, the number density of galaxies fainter than $$M=-17$$ doesn't increase that fast (in astronomish: $$-17$$ is close to $$M^*$$; "M-star").

So for the sake of this calculation, let's assume that the closest galaxy is an $$M=-17$$ galaxy at a distance of $$13\,\mathrm{Mlyr}$$. That distance corresponds to a distance modulus of $$\mu \simeq 28$$, so the apparent magnitude of the galaxy would be $$m = M + \mu \simeq 11.$$

Typically, humans cannot see objects darker than $$m \simeq 6.5$$ (the magnitude scale is backwards, so darker means "larger values than 6.5"), although some have claimed to be able to see $$m\simeq8$$ — still an order of magnitude brighter than the $$m=11$$ estimated above. Moreover, this threshold assumes point sources, whereas a galaxy has its brightness smeared out over a quite large area, lowering its surface brightness significantly!$$^\dagger$$. Note also that, as in the rest of the Universe, galaxies in voids are not completely randomly scattered throughout space, but tend to cluster in clumps and filaments, and that the number density is smaller in the center of the void, meaning that here the typical distance to the next galaxy is larger.

Hence, you would — at a random position in the Böoted Void — be most likely to be floating in complete darkness.$$^\ddagger$$

The combined light from all astrophysical and cosmological sources comprises a cosmic background radiation (CBR), meaning that at any time your eye does indeed receive photons across the entire electromagnetic spectrum. Thus the term "complete darkness" may be debated. On average, this background is dominated by the cosmic microwave background (if you're close to a star or a galaxy, those sources will dominate, but then it isn't really a "background" any longer).

In this answer, I estimate the total background in the visible region (from sources outside the Milky Way) to be roughly $$3.6\times10^{-8}\,\mathrm{W}\,\mathrm{m}^{-2}$$. If I've done my maths right, this corresponds to a the light from a 25 W light bulb, smeared out over a 15 km diameter sphere with you in the center. The Böotes Void would have an even lower background than this. I'm not a physiologist, but I think this qualifies as "complete darkness" (to the human eye; not to a telescope).

$$^\dagger$$For instance, the Andromeda galaxy has an apparent magnitude of $$m=3.44$$ which, if its light were concentrated in a point, would make it easily visibly even under light-polluted conditions.

$$^\ddagger$$Your eye might be able to detect individual photons, at stated in Árpád Szendrei's answer, but that hardly counts as "seeing anything".

• @PaulYoung Yes, anywhere in space there is a "cosmic background radiation", which is the combined field of all photons of cosmological and astrophysical origin. On average, it is dominated by the cosmic microwave background (see this answer for an overview), but if you're close to a galaxy, a star, or a flashlight, those sources will dominate. On average, optical photons (mostly from stars, and hence galaxies), is several orders of magnitudes less than the CMB, and in a void, it will be correspondingly smaller.
– pela
Apr 20, 2019 at 19:11
• So: Yes, there is light, but none detectable by a human eye.
– pela
Apr 20, 2019 at 19:12
• @PaulYoung Good point! I edited the answer to discussing the background. I also modified the claim of "complete darkness", since 1) it is only on average (a random position could place you close enough to a galaxy) and 2) the background might give the eye a different sensation than "complete darkness", but as an astronomer I cannot answer this; you might want to ask a physiologist.
– pela
Apr 22, 2019 at 8:46
• @PaulYoung Wrt. superclusters and clusters of galaxies, those are absolutely not visible to the naked eye from Earth. They are too distant, and too smeared-out. Only four galaxies (Andromeda, LMC, SMC, Triangulum) are generally visibly to the naked eye from Earth, though there has been claims from a few people that three more (Centaurus A, M81, and M83) are visible.
– pela
Apr 22, 2019 at 8:50
• @pela - that quantitative estimate really closes the coffin on the case. A 25W bulb's light smeared out over a 15km radius sphere ... Apr 23, 2019 at 14:42

According to astronomer Greg Aldering, the scale of the void is such that "If the Milky Way had been in the center of the Boötes void, we wouldn't have known there were other galaxies until the 1960s."

(The original article can be found thanks to the Wayback Machine.)

This means that if we were in the center of the Boötes void, we wouldn't be able to see any galaxies with the naked eye, and not even with some telescopes. Again according to Wikipedia, telescopes were invented at the beginning of the 17th century, radio telescopes in the 1930s and infrared telescopes in the 1960s.

In fact you would see only individual stars. Like on Earth, you can only see individual stars, and no galaxies with the naked eye. Why? Because galaxies are too dim as a whole.

Now, if you would just float in empty space in the void you would be better off then on Earth. It is because here on Earth there is some light pollution.

But in the void, there is basically no light pollution, so you could see even single photons arriving from far away giant stars.

And yes, the human eye is able to detect even single photons.

By the way, all the stars that you can see with the naked eye from Earth are from the Milky Way. But again, that is too because of the light pollution and because our night sky even if there would be no light pollution here on Earth would be filled with brighter start from the Milky way. Our eyes would get used to that, and the start from other galaxies would just appear too dim to see.

Now on the dark side of the moon it would be a little different, but only that there is no light pollution like on Earth. Still, from the dark side of the Moon you could still only see stars from the Milky way.

Now in the void in your case, I think it is different. It is so dark and void, that even a single photon coming from a far away large star would be visible to our eyes that get used to the darkness.

Do all the individual stars that we can see in the night sky belong to Milky Way?

• You can see Andromeda with the naked eye, under optimal conditions. It's mag +3.44. Sagittarius Dwarf Spheroidal Galaxy and M33 should also be visible
– CSM
Apr 17, 2019 at 20:04
• the human eye is able to detect even single photons Sort of. A human rod cell can be activated by a single photon. But the retinal circuitry extinguishes the noise of such isolated "pixels" as part of its pre-processing. However, the frog retina doesn't contain that neural circuitry, so frog brains can detect single photon events. Apr 18, 2019 at 1:06
• It depends on the number of photons per second (obviously). You can observe light sources which result in as few as one or two photons per second hitting the same rod, so I think that counts as single photons. You're right that a single photon hitting the retina randomly will not be visible. Apr 18, 2019 at 9:28
• The first sentence is completely wrong. As mentioned by CSM, you can see Andromeda Galaxy (one trillion stars), the Magellanic clouds (30 + 3 billion stars) and many clusters (e.g. Great Hercules Cluster with 300000 stars and Omega Centauri with 10 million stars) perfectly well with the naked eye. It's hard to write a credible answer after such an intro. Apr 18, 2019 at 13:41
• Could you please edit your answer? The question was featured on hot network questions. Apr 19, 2019 at 21:30

It seems that even supernovas and quasars would not be bright enough to be seen from 100 million light-years away (Boötes Void's diameter is 330 million light-years).

If you see something that looks like a star from the Boötes Void, it could be a Gamma-ray burst.

This one was visible from Earth with the naked eye (mag. 5.8) even though it was 7.5 billion light-years away.

I don't think it's likely you would see anything, considering the fact it's very dark and the human eye can normally see in the dark. I do think it's a good question though.