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It stands to reason: if it didn't, the entire sky would be covered with stars shining blindingly day and night.

But what causes a light wave (or an electromagnetic wave) to fade if there are no celestial bodies in its way? Or, alternatively, why doesn't it fade sooner? Some stars are visible to the naked eye that are hundreds of light years away. Others, closer, can only be detected with the aid of a telescope. Why?

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As the individual photons move away from a point they spread out so the same size telescope or eyeball picks up less of them as you move farther away.

There are more than two phenomena at work here (sorry HDE), in addition to just spreading out based on their initial direction individual photons can be 1. diffracted by moving into more dense space 2. reflected (by any object with a greater diameter than it's wavelength) or 3. absorbed... Especially after picking on him I must encourage everyone to upvote HDE for having a cool picture though!

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There are two phenomena at work here: the inverse-square law and extinction.

Picture a sphere centered on a star. At time $t_0$, a star emits $n$ photons, spread in random directions. At some time $t_1$, the photons are arranged in a sphere of radius $R_1$1. At some time $t_2$, the photons are arranged in a sphere of radius $R_2$, where $R_2>R_1$. This means that the photons are spread out more - and thus, the star appears dimmer at $R_2$ than it does at $R_1$. This is a consequence of the inverse-square law, because the surface area of a sphere is $4\pi r^2$.

Astronomical extinction is the absorption of photons by gas and dust spread throughout space. This can make observations difficult, because gas clouds and nebulae can easily block out most of a star's light.

In really extreme cases, a nebula or a related object will block out all the stars behind it. See, for example, Barnard 68, a molecular cloud:


Image courtesy of Wikipedia user Huntster under the Creative Commons Attribution 4.0 International license.


1 Specifically, at time $t_n$, $R_n=t_nc$, where $c$ is the speed of light.

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  • $\begingroup$ Okay, let me just make sure. In order for light to be visible, more than one photons have to hit the eye simultaneously; the distance between them has to be greater than zero yet less than X, where X is the width of eye surface responsible for registering signals. Am I right? $\endgroup$ – Ricky Jan 3 '16 at 2:51
  • $\begingroup$ @Ricky I couldn't tell you the required sensitivity (i.e. necessary number of photons). $\endgroup$ – HDE 226868 Jan 3 '16 at 3:08
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See the wikipedia article on Obler's paradox.

Basically, dust and other stuff cannot stop light for getting to us as an infinite sky filled with stars (which was dogma about a century ago) will eventually make even dark dust glow brightly.

The answer to your question is that light does not 'fade as it travels'.

Light from very distant stars is red - shifted - which I would not classify as 'fading'. So that makes distant things look dimmer.

The real answer though is that the Universe is finite in both age and size (and expanding) so that we don't see Obler's paradox.

https://en.wikipedia.org/wiki/Olbers%27_paradox

Our Sun is a star. Stars are not all the same size and brightness. There are stars a million times brighter than our Sun, and a million times dimmer. So the bright ones show up from farther away. Astronomers had to account for being able to see brighter stars more easily when they started to calculate the abundance of different types of stars in the galaxy.

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  • $\begingroup$ I beg to differ. According to your logic, EVERY SINGLE star in the universe should be visible, its actual brightness unaffected by distance. In other words, the stars in remote galaxies should be just as visible to the naked eye as Rigel, and some of them should be even brighter than Rigel, with a bit of red tint added to the view for style. $\endgroup$ – Ricky Jan 7 '16 at 0:49

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