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Light moves along straight lines... Got it.

So if the light from that faraway star is traveling in a straight line, and that beam is, considering the distaces involved, at best one or two photons "wide", why, when I move six feet or so to my left, can I still see the star? IOW, why aren't there "dark" areas between the beams of starlight, where the stars "disappear"?

Hep
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    Lots and lots of photons – anon01 Oct 28 '15 at 04:49
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    The star is not just emitting photons in one direction. It emits millions of photons in all directions. Therefore, we can see it in along varying lines of sight. – Prahar Oct 28 '15 at 04:57
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    Also, BTW, light moves in a straight line as long as the medium that is travelling in does not change. If it does, it reflects or refracts. The atmosphere of the earth causes lots of refraction so that light goes out in all directions. Thus, even if for some crazy reason, the photons were emitted along a single direction, they would be diffused by the atmosphere and give the same effect. However, note that in this case, the star would appear brighter or dimmer based on the angle that we view it in. – Prahar Oct 28 '15 at 04:57
  • But the photons are all being emitted from a single source,ie, the star, even if the photons are being scattered in all directions from the source, only those few that are traveling directly toward us are visible. they are not being generated from somewhere next to my eye – Hep Oct 28 '15 at 05:09
  • Albeit that the atmosphere does distort the rays, when we point Hubble at a star the light doesn't disappear as the telescope moves. – Hep Oct 28 '15 at 05:11
  • This is an interesting question I've wondered about. Consider Betelgeuse (D=342.5 LY, r=8.2e8 km) A sphere of radius 342.5 LY has a surface of 1.3e44 mm^2. Assuming we need say 40 photons/sec to see it, and a pupil diameter of 1 mm^2, the photon flux at the surface of the star has to be a total of 5.28e45 photons/sec or 6.45e30 photons/sec/mm^2. Is that possible? Even if my pupil diameter estimate is small by a factor of 5 it doesn't change the order of magnitude of the numbers. – Jim Garrison Oct 28 '15 at 06:22
  • Let me see if I grasp what you're saying... the biggest star out there would have to generate absurd amounts of light at its surface to be visible at all? – Hep Oct 28 '15 at 06:53
  • It's a good question, but it's been asked & answered before. – Kyle Kanos Oct 28 '15 at 10:38

2 Answers2

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I will try to address 2 points here:

1) Light does not necessarily move by rectilinear (straight line) propagation. is one example. If you also consider , then due to the curvature of space-time between 2 massive bodies, light will have to follow a curved path, known as a in transit.

2) The light from a star is not 1 beam wide, in fact, it has complete angular distribution from a celestial horizon to another. If you can't see a star at night, it may be due to its excessive dimness or due to it being beyond the horizon (notwithstanding atmospheric conditions). This very large angular distribution of its radiations allows you to see a star (don't forget that a star is in fact a sphere, not a point source of light).

Tamoghna Chowdhury
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    Got it, a star is a spheroid. but let's start from the star... beams of photons radiate outward along rays... this particular ray points directly at us... the ray next to it heads off at a slightly different angle... the farther we get from the star, the wider the gap (vector space?) between the two rays becomes... by the time it gets here, the gap is huge... so why can't I step "out of path" of that ray? – Hep Oct 28 '15 at 06:45
  • I get what you mean, but since a photon is almost dimensionless, there can be an arbitrary number of them in any sub-angle you may bother to think of, although they might not all be visible (intensity considerations). – Tamoghna Chowdhury Oct 28 '15 at 06:49
  • So photons are so tiny I'd have to be something akin to an atom to "step out" of their shine, is that the gist? – Hep Oct 28 '15 at 07:44
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    @Hep: its a mistake to think of a photon purely as if it were a particle with an exact size and location. – RedGrittyBrick Oct 28 '15 at 10:25
  • That's essentially what I'm thinking, that the reason I can always see the star has something more to do with the principles at work in the double slit experiment, ie, that until observed, the light is not 'restrained' to a straight line, hence I see the light from various positions. – Hep Oct 28 '15 at 22:00
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    It don't think this has anything to do with a double slit experiment. You can't see a photon and step out of the the way even if you are an atom. Once you have seen the photon you have already absorbed it, and having seen the first photon you still don't know where the second photon is so you can't step out of (or into) its way either. – gmatht Apr 17 '19 at 15:10
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There, in fact can be, and thus are, dark areas which causes the light to disappear.

Feynman explains this in the link; The punch line is at 4:20. But I doubt you can understand it without looking the whole series of these lectures. This must be one reason why the stars are twinkling (or scintillating)

Of course one problem is that your eyes' diameter is too big to be able to "see" these dark areas completely. The other is that these are all present in various colours.

Tamoghna Chowdhury
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Jokela
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  • Are you referring to a star's absorption spectrum? – Tamoghna Chowdhury Oct 28 '15 at 07:50
  • at 4 min 20 feynman is talking about reflection and interference between two surfaces, not a light source itself – anna v Oct 28 '15 at 09:51
  • @annav Yes, and As long as i know, the space is not a vacuum, and the direction of reflection is not very limited. Of course this aspect is very theoretical. In the praxis the different wavelengths make's this effect visible. But I think that some other kind of answer would only limit our thoughts – Jokela Oct 28 '15 at 10:35
  • What you addressed is a physical occurence and a valid point, but I don't think it was what the OP intended his question to be understood as. – Tamoghna Chowdhury Oct 29 '15 at 16:37
  • Yes. Light is constructed from single photons. And there is a distance between them. And their path is a straight line only in vacuum. As we are in Physics-platform, I think it's not proper to make same un-true simplifications, which would only disturb further learning; https://www.youtube.com/watch?v=iMDTcMD6pOw – Jokela Nov 03 '15 at 10:39