<p>What's more, stars are so small to our eyes that they look like "point sources," meaning that each resolution element in our eyes gathers light from a much larger region of the sky than a distant star can fill.  If you were to look at the Sun from, say, Uranus, and you have good eyes, it would still hurt almost as much because you could still resolve the Sun at that distance (sharp eyes can resolve objects 1/25 the width of the Sun).  That means whatever spot on your optic nerve the sunlight is focused to would be getting the same heat that it gets now-- there would be fewer such spots if you were at Uranus, but a burning spot still hurts.  </p>

<p>But any distance larger than 25 astronomical units or so, you would start to not be able to resolve the Sun so even that single spot on the optic nerve would be mixing in light from the dark region around the Sun.  At <span class="math-container">$10^4$</span> times farther away than Uranus, the distance mentioned in the other answer, it means the sunlight would get diluted by the darkness around the Sun <span class="math-container">$10^8$</span> times larger than the Sun, accordingly reducing the heat that spot on the optic nerve receives by <span class="math-container">$10^{-8}$</span>.  Far enough away, you can't even see there is a star there.</p>

CC BY-SA 4.0
2019-05-20T10:40:00.953Z
Mixing in light coming from dark is a bit bizarre of a formulation. It does suggest dark rays coming to the eyes instead of light rays diluting to space. A very curious explanation of the inverse square law and related flux. What would happen with a fluid?
Alchimista
CC BY-SA 4.0
2019-05-20T17:00:32.243Z
There's always some light, even if only CMB.  The key point is, there are really two very separate types of inverse-square behavior, one where the source is resolved and its specific intensity stays fixed but the solid angle it subtends drops by inverse-square, and the other is when the source is not resolved and it appears to get dimmer but not smaller.  It's mostly that latter type that controls damage to the optic nerve.  I don't know what you mean about a fluid though.
Ken G
CC BY-SA 4.0
2019-05-21T07:09:27.853Z
It is fine to bring the aspect of resolution in the answer. The answer fails with such sentences " means the sunlight would get diluted by the darkness around the Sun  times larger than the Sun, accordingly reducing the heat that spot on the optic nerve receives by . Far enough away, you can't even see there is a star there." Nothing dilutes the light from over there. The light is diluted by dustance and that is.As for a fluid coming from a central point. It is a wrong way to illustrate the situation. As soon the light can damage a eye spot, it does, no matter how much dark is visible.
Alchimista
CC BY-SA 4.0
2019-05-22T13:44:22.027Z
Well, you simply didn't understand the entire point.  Distance by itself is not the determining factor to how much it hurts your eye, brightness on each resolution element in the eye is.  Distance doesn't affect that until you get quite far away.   If you simply consider a single resolution element on your optic nerve, you find no change to very large distance, followed by a reduction due to the fact that the resolution element "sees" a combination of bright rays and dark rays.  So something does dilute the brightness of the bright rays-- the dark rays.  And yes, a ray can be dark.
Ken G
CC BY-SA 4.0
2019-05-22T14:16:54.397Z
The problem with your interpretation is probably that you are regarding a "ray" as a physical thing, and indeed sometimes the word is used that way.  But mathematically, a "ray" is a concept, an imagined straight line.  And when interpreting the brightness of an observation, one can equip each imagined ray with a quantity called "specific intensity", which can be bright or dark depending on what source illuminates each ray.  This is the technique known as "ray tracing" to determine the brightness of an observation.
Ken G
CC BY-SA 4.0
2019-05-22T16:42:57.983Z
Anyway I was specifically motivated by that amended sentencr and anyway, because you are insisting, there is not dilution still. If the photons impinging the receptor are enough it will be damaged no matter how much dark is in your scope @Ken G.
Alchimista
CC BY-SA 4.0
2019-05-15T15:42:05.400Z
2019-05-15T18:22:11.950Z
2019-05-15T18:22:11.950Z
Ken G
<p>The sun is one "astronomic unit" away. The unit is defined that way. The closest other star is <a href="https://imagine.gsfc.nasa.gov/features/cosmic/nearest_star_info.html" rel="nofollow noreferrer">Proxima Centauri</a> at a distance of 268,770 astronomic units away. The intensity of the light reduces at distance squared; if Proxima Centauri sent out just as much light as the sun we would see it as (268,770)<sup>2</sup> times fainter. That is approximately 72 billion times fainter. Different stars vary in their luminosity but they are all a long way away, except for the Sun. That is why you can look at them without hurting your eyes.</p>

CC BY-SA 4.0
2019-05-15T15:22:47.553Z
2019-05-15T18:22:22.720Z
2019-05-15T18:22:22.720Z
Agerhell
<p>Distance <em>is</em> important, as is surface brightness, but there's also one more thing that has not been let onto here.</p>

<p>First, a recap. For one, the fact that while <em>total</em> brightness falls with distance, <em>surface</em> brightness does not is indeed that - an unassailable fact, as simply observing whether or not the surface brightness of a wall changes as you move closer to, or further from, it. (It doesn't.)</p>

<p>So, if you get twice as far from the Sun (e.g. 300 Gm versus 150 Gm in the case of "as far" being Earth's distance), then while its angular size halves and area quarters, as you might expect, that quarter-sized area has just as much brightness - just as much "punch", if you will - as the full-size one at our distance and thus just as much is capable of burning (that much smaller) a hole in your retina.</p>

<p>When you get to the distance of the nearest stars at Proxima and Alpha Centauri, that is 40 Pm distant, or <span class="math-container">$40 \times 10^6\ \mathrm{Gm}$</span>. Centauri provides a reasonable test case since it is a binary system of two stars that are roughly similar to the Sun. Thus, we can roughstimate it as being like a Sun some <span class="math-container">$\frac{40\ 000\ 000}{150}$</span> or about 270,000 times, further away. That makes its angular size on the sky, and thus actual size on our retina - about 1/270,000 that of the Sun. If the spot on the retina is on the order of 1 mm from the ordinary Sun (say 0.1-1 - don't know offhand), then this will be in the range of <span class="math-container">$10^{-6}\ \mathrm{mm}$</span>, i.e. the single-digit <em>nanometer</em> range, at best.</p>

<p>In theory, such a spot could still perhaps get a few atoms plenty hot depending on just where in that range it falls and could, say, perhaps puncture a cell membrane, but the trick is, in fact, you never actually <em>can</em> get a spot in single-digit nanometers with visible light, much less near-infrared, because of <em>diffraction</em>: the wavelengths of visible light are at about 400-750 nm, and that sets the scale of the largest "spot" possible since a full wave cycle could not fit in any smaller an area. Effectively, the distant star is <em>not</em> a tiny spot, even with perfect focus, but actually a tiny <em>blur</em> on your retina and even then, I believe many, if not all, human eyes don't even reach the diffraction limit to begin with (<em>esp.</em> if you're myopic!), making it even <em>more</em> blurred. At even the low end of this range, 400 nm, the surface brightness now <em>is</em> diluted because it's the same total light - that entered the eyeball - spread over a larger area, on the order of <span class="math-container">$400^2$</span> or <em>16,000</em> times fainter <em>per unit area</em>. Clearly, that is not going to injure anything <em>at all</em>.</p>

CC BY-SA 4.0
2019-05-16T23:31:45.960Z
2019-05-17T12:17:56.900Z
2019-05-17T12:17:56.900Z
The_Sympathizer
<p>If I stare at the sun, it's bright and blinding. However, if I look at a star in the sky, it doesn't hurt. <strong>What causes this loss of energy that would otherwise damage one's eyes?</strong></p>

CC BY-SA 4.0
2019-05-15T15:13:37.917Z
In a word, distance. See https://en.wikipedia.org/wiki/Inverse-square_law
PM 2Ring
<star><light>
Why doesn't it hurt to look at stars?
Stormblessed
# Why doesn't it hurt to look at stars?

(Asked by: username_0 on 2019-05-15)

If I stare at the sun, it's bright and blinding. However, if I look at a star in the sky, it doesn't hurt. **What causes this loss of energy that would otherwise damage one's eyes?**

**Tags:** star, light
> **Comments:**
> **username_1**: In a word, distance. See https://en.wikipedia.org/wiki/Inverse-square_law  

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## 3 answers

### Answer by username_2 on 2019-05-15. Score: 4

What's more, stars are so small to our eyes that they look like "point sources," meaning that each resolution element in our eyes gathers light from a much larger region of the sky than a distant star can fill. If you were to look at the Sun from, say, Uranus, and you have good eyes, it would still hurt almost as much because you could still resolve the Sun at that distance (sharp eyes can resolve objects 1/25 the width of the Sun). That means whatever spot on your optic nerve the sunlight is focused to would be getting the same heat that it gets now-- there would be fewer such spots if you were at Uranus, but a burning spot still hurts.

But any distance larger than 25 astronomical units or so, you would start to not be able to resolve the Sun so even that single spot on the optic nerve would be mixing in light from the dark region around the Sun. At $10^4$ times farther away than Uranus, the distance mentioned in the other answer, it means the sunlight would get diluted by the darkness around the Sun $10^8$ times larger than the Sun, accordingly reducing the heat that spot on the optic nerve receives by $10^{-8}$. Far enough away, you can't even see there is a star there.

> **Comments:**
> **username_3**: Mixing in light coming from dark is a bit bizarre of a formulation. It does suggest dark rays coming to the eyes instead of light rays diluting to space. A very curious explanation of the inverse square law and related flux. What would happen with a fluid?  
> **username_2**: There's always some light, even if only CMB.  The key point is, there are really two very separate types of inverse-square behavior, one where the source is resolved and its specific intensity stays fixed but the solid angle it subtends drops by inverse-square, and the other is when the source is not resolved and it appears to get dimmer but not smaller.  It's mostly that latter type that controls damage to the optic nerve.  I don't know what you mean about a fluid though.  
> **username_3**: It is fine to bring the aspect of resolution in the answer. The answer fails with such sentences " means the sunlight would get diluted by the darkness around the Sun  times larger than the Sun, accordingly reducing the heat that spot on the optic nerve receives by . Far enough away, you can't even see there is a star there." Nothing dilutes the light from over there. The light is diluted by dustance and that is.As for a fluid coming from a central point. It is a wrong way to illustrate the situation. As soon the light can damage a eye spot, it does, no matter how much dark is visible.  
> **username_2**: Well, you simply didn't understand the entire point.  Distance by itself is not the determining factor to how much it hurts your eye, brightness on each resolution element in the eye is.  Distance doesn't affect that until you get quite far away.   If you simply consider a single resolution element on your optic nerve, you find no change to very large distance, followed by a reduction due to the fact that the resolution element "sees" a combination of bright rays and dark rays.  So something does dilute the brightness of the bright rays-- the dark rays.  And yes, a ray can be dark.  
> **username_2**: The problem with your interpretation is probably that you are regarding a "ray" as a physical thing, and indeed sometimes the word is used that way.  But mathematically, a "ray" is a concept, an imagined straight line.  And when interpreting the brightness of an observation, one can equip each imagined ray with a quantity called "specific intensity", which can be bright or dark depending on what source illuminates each ray.  This is the technique known as "ray tracing" to determine the brightness of an observation.  
> **username_3**: Anyway I was specifically motivated by that amended sentencr and anyway, because you are insisting, there is not dilution still. If the photons impinging the receptor are enough it will be damaged no matter how much dark is in your scope @Ken G.  

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### Answer by username_4 on 2019-05-15. Score: 2

The sun is one "astronomic unit" away. The unit is defined that way. The closest other star is [Proxima Centauri](https://imagine.gsfc.nasa.gov/features/cosmic/nearest_star_info.html) at a distance of 268,770 astronomic units away. The intensity of the light reduces at distance squared; if Proxima Centauri sent out just as much light as the sun we would see it as (268,770)2 times fainter. That is approximately 72 billion times fainter. Different stars vary in their luminosity but they are all a long way away, except for the Sun. That is why you can look at them without hurting your eyes.


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### Answer by username_5 on 2019-05-16. Score: 2

Distance *is* important, as is surface brightness, but there's also one more thing that has not been let onto here.

First, a recap. For one, the fact that while *total* brightness falls with distance, *surface* brightness does not is indeed that - an unassailable fact, as simply observing whether or not the surface brightness of a wall changes as you move closer to, or further from, it. (It doesn't.)

So, if you get twice as far from the Sun (e.g. 300 Gm versus 150 Gm in the case of "as far" being Earth's distance), then while its angular size halves and area quarters, as you might expect, that quarter-sized area has just as much brightness - just as much "punch", if you will - as the full-size one at our distance and thus just as much is capable of burning (that much smaller) a hole in your retina.

When you get to the distance of the nearest stars at Proxima and Alpha Centauri, that is 40 Pm distant, or $40 \times 10^6\ \mathrm{Gm}$. Centauri provides a reasonable test case since it is a binary system of two stars that are roughly similar to the Sun. Thus, we can roughstimate it as being like a Sun some $\frac{40\ 000\ 000}{150}$ or about 270,000 times, further away. That makes its angular size on the sky, and thus actual size on our retina - about 1/270,000 that of the Sun. If the spot on the retina is on the order of 1 mm from the ordinary Sun (say 0.1-1 - don't know offhand), then this will be in the range of $10^{-6}\ \mathrm{mm}$, i.e. the single-digit *nanometer* range, at best.

In theory, such a spot could still perhaps get a few atoms plenty hot depending on just where in that range it falls and could, say, perhaps puncture a cell membrane, but the trick is, in fact, you never actually *can* get a spot in single-digit nanometers with visible light, much less near-infrared, because of *diffraction*: the wavelengths of visible light are at about 400-750 nm, and that sets the scale of the largest "spot" possible since a full wave cycle could not fit in any smaller an area. Effectively, the distant star is *not* a tiny spot, even with perfect focus, but actually a tiny *blur* on your retina and even then, I believe many, if not all, human eyes don't even reach the diffraction limit to begin with (*esp.* if you're myopic!), making it even *more* blurred. At even the low end of this range, 400 nm, the surface brightness now *is* diluted because it's the same total light - that entered the eyeball - spread over a larger area, on the order of $400^2$ or *16,000* times fainter *per unit area*. Clearly, that is not going to injure anything *at all*.


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