Can anything be seen from the center of the Boötes void? How dark would it be?Do all the individual stars that we can see in the night sky belong to Milky Way?Does the number density of photons $n_gammaapprox 10^8 :mathrm m^-3$ refer to CMB photons only?How bright is the night sky in $fracmathrmWmathrmm^2$?What does the sky look like to human eyes from orbit?Names, maps for Milky Way dust clouds? (Dark Cloud Constellations)What is the farthest-away star visible to the naked eye?View of the Sun from Voyager 1How deep is the Great Red Spot?Does the Milky Way have dark matter satellite galaxies?Establishing true (ie present) positions of cosmic bodiesIf someone from another part of the observable universe had their own Hubble Telescope, would they see different things than we did?
Why do English transliterations of Arabic names have so many Qs in them?
How do lasers measure short distances (<1cm) when electronics are too slow for time-of-flight to work?
Can a Dragon enter the Feywild without having to use fey crossing portals?
What is the next number in the series: 21, 21, 23, 20, 5, 25, 31, 24,?
Can I color text by using an image, so that the color isn't flat?
Why are engines with carburetors hard to start in cold weather?
Find the percentage
Why can I ping 10.0.0.0/8 addresses from a 192.168.1.0/24 subnet?
How do I break the broom in Untitled Goose Game?
How to find out which object is taking space?
Can massive damage kill you while at 0 HP?
one-liner vs script
Relation between signal processing and control systems engineering?
Why is the logical NOT operator in C-style languages "!" and not "~~"?
Are there any privately owned large commercial airports?
Can I perform Umrah while on a Saudi Arabian visit e-visa
How can I change the MSX function keys using ASM?
Overlay image with parts of another image
Lazy Approval In SharePoint Online
My first random password generator
div innerHTML with standard lightning web components
How to make a gift without seeming creepy?
What do you call the fallacy of thinking that some action A will guarantee some outcome B, when in reality B depends on multiple other conditions?
how do you value what your leisure time is worth?
Can anything be seen from the center of the Boötes void? How dark would it be?
Do all the individual stars that we can see in the night sky belong to Milky Way?Does the number density of photons $n_gammaapprox 10^8 :mathrm m^-3$ refer to CMB photons only?How bright is the night sky in $fracmathrmWmathrmm^2$?What does the sky look like to human eyes from orbit?Names, maps for Milky Way dust clouds? (Dark Cloud Constellations)What is the farthest-away star visible to the naked eye?View of the Sun from Voyager 1How deep is the Great Red Spot?Does the Milky Way have dark matter satellite galaxies?Establishing true (ie present) positions of cosmic bodiesIf someone from another part of the observable universe had their own Hubble Telescope, would they see different things than we did?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
$begingroup$
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.
astronomy
$endgroup$
|
show 5 more comments
$begingroup$
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.
astronomy
$endgroup$
$begingroup$
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.
$endgroup$
– JMac
Apr 17 at 14:29
5
$begingroup$
Let's pick the spot that is furthest from any galaxy.
$endgroup$
– Paul Young
Apr 17 at 14:32
11
$begingroup$
@PaulYoung so you're saying "if there's a bright center to the Universe, you're at the [point] it's farthest from"?
$endgroup$
– MikeTheLiar
Apr 17 at 21:47
3
$begingroup$
@MikeTheLiar seems more like "let's get as far away from as many galaxies as possible"
$endgroup$
– immibis
Apr 18 at 0:38
$begingroup$
As a starter: What would we see from Earth if we removed all those celestial objects that belong to our Galaxy?
$endgroup$
– Hagen von Eitzen
Apr 19 at 10:05
|
show 5 more comments
$begingroup$
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.
astronomy
$endgroup$
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.
astronomy
astronomy
edited Apr 17 at 20:31
costrom
3141 gold badge2 silver badges8 bronze badges
3141 gold badge2 silver badges8 bronze badges
asked Apr 17 at 14:24
Paul YoungPaul Young
2,4408 silver badges27 bronze badges
2,4408 silver badges27 bronze badges
$begingroup$
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.
$endgroup$
– JMac
Apr 17 at 14:29
5
$begingroup$
Let's pick the spot that is furthest from any galaxy.
$endgroup$
– Paul Young
Apr 17 at 14:32
11
$begingroup$
@PaulYoung so you're saying "if there's a bright center to the Universe, you're at the [point] it's farthest from"?
$endgroup$
– MikeTheLiar
Apr 17 at 21:47
3
$begingroup$
@MikeTheLiar seems more like "let's get as far away from as many galaxies as possible"
$endgroup$
– immibis
Apr 18 at 0:38
$begingroup$
As a starter: What would we see from Earth if we removed all those celestial objects that belong to our Galaxy?
$endgroup$
– Hagen von Eitzen
Apr 19 at 10:05
|
show 5 more comments
$begingroup$
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.
$endgroup$
– JMac
Apr 17 at 14:29
5
$begingroup$
Let's pick the spot that is furthest from any galaxy.
$endgroup$
– Paul Young
Apr 17 at 14:32
11
$begingroup$
@PaulYoung so you're saying "if there's a bright center to the Universe, you're at the [point] it's farthest from"?
$endgroup$
– MikeTheLiar
Apr 17 at 21:47
3
$begingroup$
@MikeTheLiar seems more like "let's get as far away from as many galaxies as possible"
$endgroup$
– immibis
Apr 18 at 0:38
$begingroup$
As a starter: What would we see from Earth if we removed all those celestial objects that belong to our Galaxy?
$endgroup$
– Hagen von Eitzen
Apr 19 at 10:05
$begingroup$
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.
$endgroup$
– JMac
Apr 17 at 14:29
$begingroup$
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.
$endgroup$
– JMac
Apr 17 at 14:29
5
5
$begingroup$
Let's pick the spot that is furthest from any galaxy.
$endgroup$
– Paul Young
Apr 17 at 14:32
$begingroup$
Let's pick the spot that is furthest from any galaxy.
$endgroup$
– Paul Young
Apr 17 at 14:32
11
11
$begingroup$
@PaulYoung so you're saying "if there's a bright center to the Universe, you're at the [point] it's farthest from"?
$endgroup$
– MikeTheLiar
Apr 17 at 21:47
$begingroup$
@PaulYoung so you're saying "if there's a bright center to the Universe, you're at the [point] it's farthest from"?
$endgroup$
– MikeTheLiar
Apr 17 at 21:47
3
3
$begingroup$
@MikeTheLiar seems more like "let's get as far away from as many galaxies as possible"
$endgroup$
– immibis
Apr 18 at 0:38
$begingroup$
@MikeTheLiar seems more like "let's get as far away from as many galaxies as possible"
$endgroup$
– immibis
Apr 18 at 0:38
$begingroup$
As a starter: What would we see from Earth if we removed all those celestial objects that belong to our Galaxy?
$endgroup$
– Hagen von Eitzen
Apr 19 at 10:05
$begingroup$
As a starter: What would we see from Earth if we removed all those celestial objects that belong to our Galaxy?
$endgroup$
– Hagen von Eitzen
Apr 19 at 10:05
|
show 5 more comments
5 Answers
5
active
oldest
votes
$begingroup$
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 $n sim 0.004,mathrmMpc^-3$, or $10^-4,mathrmMlyr^-3$ (i.e. "per cubic mega-light-year"). Hence, the typical distance to a galaxy from a random point in the Boötes Void is
$$
d = left( frac34pi n right)^1/3 simeq 13,mathrmMlyr.
$$
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 $13,mathrmMlyr$. 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 $msimeq8$ — 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.$^ddagger$
Background radiation
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 $3.6times10^-8,mathrmW,mathrmm^-2$. 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".
$endgroup$
$begingroup$
I agree with your conclusion, but I would not agree with your arguments. Yes, one will not be able to see individual galaxies. But that doesn't mean that there will be total darkness. One can't see individual stars forming the Milky Way, but still can see the bright area on the sky.
$endgroup$
– lesnik
Apr 19 at 13:26
1
$begingroup$
So: Yes, there is light, but none detectable by a human eye.
$endgroup$
– pela
Apr 20 at 19:12
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:46
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:50
1
$begingroup$
@pela - that quantitative estimate really closes the coffin on the case. A 25W bulb's light smeared out over a 15km radius sphere ...
$endgroup$
– Paul Young
Apr 23 at 14:42
|
show 5 more comments
$begingroup$
From the Wikipedia page on the Boötes void (the same that you linked to in your question):
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.
$endgroup$
add a comment
|
$begingroup$
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.
Please see here:
Do all the individual stars that we can see in the night sky belong to Milky Way?
$endgroup$
13
$begingroup$
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
$endgroup$
– CSM
Apr 17 at 20:04
12
$begingroup$
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.
$endgroup$
– PM 2Ring
Apr 18 at 1:06
$begingroup$
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.
$endgroup$
– jcupitt
Apr 18 at 9:28
10
$begingroup$
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.
$endgroup$
– Eric Duminil
Apr 18 at 13:41
1
$begingroup$
Could you please edit your answer? The question was featured on hot network questions.
$endgroup$
– Eric Duminil
Apr 19 at 21:30
add a comment
|
$begingroup$
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.
$endgroup$
add a comment
|
$begingroup$
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.
$endgroup$
add a comment
|
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/4.0/"u003ecc by-sa 4.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f473330%2fcan-anything-be-seen-from-the-center-of-the-bo%25c3%25b6tes-void-how-dark-would-it-be%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $n sim 0.004,mathrmMpc^-3$, or $10^-4,mathrmMlyr^-3$ (i.e. "per cubic mega-light-year"). Hence, the typical distance to a galaxy from a random point in the Boötes Void is
$$
d = left( frac34pi n right)^1/3 simeq 13,mathrmMlyr.
$$
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 $13,mathrmMlyr$. 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 $msimeq8$ — 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.$^ddagger$
Background radiation
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 $3.6times10^-8,mathrmW,mathrmm^-2$. 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".
$endgroup$
$begingroup$
I agree with your conclusion, but I would not agree with your arguments. Yes, one will not be able to see individual galaxies. But that doesn't mean that there will be total darkness. One can't see individual stars forming the Milky Way, but still can see the bright area on the sky.
$endgroup$
– lesnik
Apr 19 at 13:26
1
$begingroup$
So: Yes, there is light, but none detectable by a human eye.
$endgroup$
– pela
Apr 20 at 19:12
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:46
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:50
1
$begingroup$
@pela - that quantitative estimate really closes the coffin on the case. A 25W bulb's light smeared out over a 15km radius sphere ...
$endgroup$
– Paul Young
Apr 23 at 14:42
|
show 5 more comments
$begingroup$
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 $n sim 0.004,mathrmMpc^-3$, or $10^-4,mathrmMlyr^-3$ (i.e. "per cubic mega-light-year"). Hence, the typical distance to a galaxy from a random point in the Boötes Void is
$$
d = left( frac34pi n right)^1/3 simeq 13,mathrmMlyr.
$$
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 $13,mathrmMlyr$. 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 $msimeq8$ — 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.$^ddagger$
Background radiation
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 $3.6times10^-8,mathrmW,mathrmm^-2$. 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".
$endgroup$
$begingroup$
I agree with your conclusion, but I would not agree with your arguments. Yes, one will not be able to see individual galaxies. But that doesn't mean that there will be total darkness. One can't see individual stars forming the Milky Way, but still can see the bright area on the sky.
$endgroup$
– lesnik
Apr 19 at 13:26
1
$begingroup$
So: Yes, there is light, but none detectable by a human eye.
$endgroup$
– pela
Apr 20 at 19:12
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:46
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:50
1
$begingroup$
@pela - that quantitative estimate really closes the coffin on the case. A 25W bulb's light smeared out over a 15km radius sphere ...
$endgroup$
– Paul Young
Apr 23 at 14:42
|
show 5 more comments
$begingroup$
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 $n sim 0.004,mathrmMpc^-3$, or $10^-4,mathrmMlyr^-3$ (i.e. "per cubic mega-light-year"). Hence, the typical distance to a galaxy from a random point in the Boötes Void is
$$
d = left( frac34pi n right)^1/3 simeq 13,mathrmMlyr.
$$
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 $13,mathrmMlyr$. 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 $msimeq8$ — 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.$^ddagger$
Background radiation
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 $3.6times10^-8,mathrmW,mathrmm^-2$. 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".
$endgroup$
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 $n sim 0.004,mathrmMpc^-3$, or $10^-4,mathrmMlyr^-3$ (i.e. "per cubic mega-light-year"). Hence, the typical distance to a galaxy from a random point in the Boötes Void is
$$
d = left( frac34pi n right)^1/3 simeq 13,mathrmMlyr.
$$
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 $13,mathrmMlyr$. 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 $msimeq8$ — 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.$^ddagger$
Background radiation
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 $3.6times10^-8,mathrmW,mathrmm^-2$. 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".
edited Apr 23 at 9:16
answered Apr 17 at 16:16
pelapela
9,02123 silver badges39 bronze badges
9,02123 silver badges39 bronze badges
$begingroup$
I agree with your conclusion, but I would not agree with your arguments. Yes, one will not be able to see individual galaxies. But that doesn't mean that there will be total darkness. One can't see individual stars forming the Milky Way, but still can see the bright area on the sky.
$endgroup$
– lesnik
Apr 19 at 13:26
1
$begingroup$
So: Yes, there is light, but none detectable by a human eye.
$endgroup$
– pela
Apr 20 at 19:12
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:46
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:50
1
$begingroup$
@pela - that quantitative estimate really closes the coffin on the case. A 25W bulb's light smeared out over a 15km radius sphere ...
$endgroup$
– Paul Young
Apr 23 at 14:42
|
show 5 more comments
$begingroup$
I agree with your conclusion, but I would not agree with your arguments. Yes, one will not be able to see individual galaxies. But that doesn't mean that there will be total darkness. One can't see individual stars forming the Milky Way, but still can see the bright area on the sky.
$endgroup$
– lesnik
Apr 19 at 13:26
1
$begingroup$
So: Yes, there is light, but none detectable by a human eye.
$endgroup$
– pela
Apr 20 at 19:12
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:46
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:50
1
$begingroup$
@pela - that quantitative estimate really closes the coffin on the case. A 25W bulb's light smeared out over a 15km radius sphere ...
$endgroup$
– Paul Young
Apr 23 at 14:42
$begingroup$
I agree with your conclusion, but I would not agree with your arguments. Yes, one will not be able to see individual galaxies. But that doesn't mean that there will be total darkness. One can't see individual stars forming the Milky Way, but still can see the bright area on the sky.
$endgroup$
– lesnik
Apr 19 at 13:26
$begingroup$
I agree with your conclusion, but I would not agree with your arguments. Yes, one will not be able to see individual galaxies. But that doesn't mean that there will be total darkness. One can't see individual stars forming the Milky Way, but still can see the bright area on the sky.
$endgroup$
– lesnik
Apr 19 at 13:26
1
1
$begingroup$
So: Yes, there is light, but none detectable by a human eye.
$endgroup$
– pela
Apr 20 at 19:12
$begingroup$
So: Yes, there is light, but none detectable by a human eye.
$endgroup$
– pela
Apr 20 at 19:12
1
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:46
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:46
1
1
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:50
$begingroup$
@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.
$endgroup$
– pela
Apr 22 at 8:50
1
1
$begingroup$
@pela - that quantitative estimate really closes the coffin on the case. A 25W bulb's light smeared out over a 15km radius sphere ...
$endgroup$
– Paul Young
Apr 23 at 14:42
$begingroup$
@pela - that quantitative estimate really closes the coffin on the case. A 25W bulb's light smeared out over a 15km radius sphere ...
$endgroup$
– Paul Young
Apr 23 at 14:42
|
show 5 more comments
$begingroup$
From the Wikipedia page on the Boötes void (the same that you linked to in your question):
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.
$endgroup$
add a comment
|
$begingroup$
From the Wikipedia page on the Boötes void (the same that you linked to in your question):
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.
$endgroup$
add a comment
|
$begingroup$
From the Wikipedia page on the Boötes void (the same that you linked to in your question):
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.
$endgroup$
From the Wikipedia page on the Boötes void (the same that you linked to in your question):
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.
answered Apr 18 at 0:50
Fabio TuratiFabio Turati
2112 silver badges6 bronze badges
2112 silver badges6 bronze badges
add a comment
|
add a comment
|
$begingroup$
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.
Please see here:
Do all the individual stars that we can see in the night sky belong to Milky Way?
$endgroup$
13
$begingroup$
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
$endgroup$
– CSM
Apr 17 at 20:04
12
$begingroup$
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.
$endgroup$
– PM 2Ring
Apr 18 at 1:06
$begingroup$
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.
$endgroup$
– jcupitt
Apr 18 at 9:28
10
$begingroup$
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.
$endgroup$
– Eric Duminil
Apr 18 at 13:41
1
$begingroup$
Could you please edit your answer? The question was featured on hot network questions.
$endgroup$
– Eric Duminil
Apr 19 at 21:30
add a comment
|
$begingroup$
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.
Please see here:
Do all the individual stars that we can see in the night sky belong to Milky Way?
$endgroup$
13
$begingroup$
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
$endgroup$
– CSM
Apr 17 at 20:04
12
$begingroup$
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.
$endgroup$
– PM 2Ring
Apr 18 at 1:06
$begingroup$
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.
$endgroup$
– jcupitt
Apr 18 at 9:28
10
$begingroup$
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.
$endgroup$
– Eric Duminil
Apr 18 at 13:41
1
$begingroup$
Could you please edit your answer? The question was featured on hot network questions.
$endgroup$
– Eric Duminil
Apr 19 at 21:30
add a comment
|
$begingroup$
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.
Please see here:
Do all the individual stars that we can see in the night sky belong to Milky Way?
$endgroup$
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.
Please see here:
Do all the individual stars that we can see in the night sky belong to Milky Way?
answered Apr 17 at 15:40
Árpád SzendreiÁrpád Szendrei
7,2931 gold badge11 silver badges35 bronze badges
7,2931 gold badge11 silver badges35 bronze badges
13
$begingroup$
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
$endgroup$
– CSM
Apr 17 at 20:04
12
$begingroup$
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.
$endgroup$
– PM 2Ring
Apr 18 at 1:06
$begingroup$
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.
$endgroup$
– jcupitt
Apr 18 at 9:28
10
$begingroup$
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.
$endgroup$
– Eric Duminil
Apr 18 at 13:41
1
$begingroup$
Could you please edit your answer? The question was featured on hot network questions.
$endgroup$
– Eric Duminil
Apr 19 at 21:30
add a comment
|
13
$begingroup$
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
$endgroup$
– CSM
Apr 17 at 20:04
12
$begingroup$
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.
$endgroup$
– PM 2Ring
Apr 18 at 1:06
$begingroup$
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.
$endgroup$
– jcupitt
Apr 18 at 9:28
10
$begingroup$
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.
$endgroup$
– Eric Duminil
Apr 18 at 13:41
1
$begingroup$
Could you please edit your answer? The question was featured on hot network questions.
$endgroup$
– Eric Duminil
Apr 19 at 21:30
13
13
$begingroup$
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
$endgroup$
– CSM
Apr 17 at 20:04
$begingroup$
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
$endgroup$
– CSM
Apr 17 at 20:04
12
12
$begingroup$
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.
$endgroup$
– PM 2Ring
Apr 18 at 1:06
$begingroup$
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.
$endgroup$
– PM 2Ring
Apr 18 at 1:06
$begingroup$
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.
$endgroup$
– jcupitt
Apr 18 at 9:28
$begingroup$
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.
$endgroup$
– jcupitt
Apr 18 at 9:28
10
10
$begingroup$
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.
$endgroup$
– Eric Duminil
Apr 18 at 13:41
$begingroup$
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.
$endgroup$
– Eric Duminil
Apr 18 at 13:41
1
1
$begingroup$
Could you please edit your answer? The question was featured on hot network questions.
$endgroup$
– Eric Duminil
Apr 19 at 21:30
$begingroup$
Could you please edit your answer? The question was featured on hot network questions.
$endgroup$
– Eric Duminil
Apr 19 at 21:30
add a comment
|
$begingroup$
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.
$endgroup$
add a comment
|
$begingroup$
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.
$endgroup$
add a comment
|
$begingroup$
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.
$endgroup$
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.
answered Apr 20 at 14:52
Eric DuminilEric Duminil
1,0586 silver badges9 bronze badges
1,0586 silver badges9 bronze badges
add a comment
|
add a comment
|
$begingroup$
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.
$endgroup$
add a comment
|
$begingroup$
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.
$endgroup$
add a comment
|
$begingroup$
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.
$endgroup$
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.
answered Apr 23 at 10:53
RooRoo
111 bronze badge
111 bronze badge
add a comment
|
add a comment
|
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f473330%2fcan-anything-be-seen-from-the-center-of-the-bo%25c3%25b6tes-void-how-dark-would-it-be%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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.
$endgroup$
– JMac
Apr 17 at 14:29
5
$begingroup$
Let's pick the spot that is furthest from any galaxy.
$endgroup$
– Paul Young
Apr 17 at 14:32
11
$begingroup$
@PaulYoung so you're saying "if there's a bright center to the Universe, you're at the [point] it's farthest from"?
$endgroup$
– MikeTheLiar
Apr 17 at 21:47
3
$begingroup$
@MikeTheLiar seems more like "let's get as far away from as many galaxies as possible"
$endgroup$
– immibis
Apr 18 at 0:38
$begingroup$
As a starter: What would we see from Earth if we removed all those celestial objects that belong to our Galaxy?
$endgroup$
– Hagen von Eitzen
Apr 19 at 10:05