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The rainbow! What magical "thing". And even if you see the droplets of rain move in a sunlit storm, she's steady. I have been trying to understand but there are so many drops involved! And they are moving in turbulent ways on top.
So what's going on? I know that each droplet sends a "rain circle" cone towards us, and that our eyes are sprayed with these cones. These cones all have the same orientation, no matter how the droplets move. Somehow this must be the key, but I don't see how.

G. Smith
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Deschele Schilder
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    Statistics????? Raindrops are small and many but the rainbow is huge by the time you can see it. – DKNguyen May 23 '21 at 21:14
  • @DKNguyen When you involve statistics, wouldn't all the contributions cancel? – Deschele Schilder May 23 '21 at 21:19
  • Only the variations that oppose each other would. – DKNguyen May 23 '21 at 21:20
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    Note also that every rainbow is centred around your head at the same angle above the anti-solar point. So if you and your friend are standing next to each other, you are each observing your own personal rainbow, from an independent set of drops. The same is true in time: the rainbow you see in this instant is created by a different set of drops from the one you see in the next instant. So there is very little static about it, the drops' only job seems to be to refract the light and move on to create someone else's rainbow ;). – Philip May 23 '21 at 21:24
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    "When you involve statistics, wouldn't all the contributions cancel?" – No, definitely not. That's a common misconception. Suppose you flip 1,000,000,000,000 coins, and you compare the number of heads to the number of tails. I think a lot of people think that the difference will be very small, something like 10 or 20. Actually, the difference will probably be on the order of 1,000,000, and it's very, very unlikely that the difference will be less than 1,000. – Tanner Swett May 24 '21 at 12:21
  • @TannerSwett But one time this 1,000,000 would be heads (more or less), one time tails(more or less). So you have to perform the 1,000,000,000,000 times experiment 1,000,000,000,000 times. The number of views is 666... – Deschele Schilder May 24 '21 at 12:40
  • Sure, so perform the experiment 1,000,000,000,000 times, and each time you get more heads than tails, write down "HEADS," and each time you get more tails than heads, write down "TAILS." Then look at the difference between the number of words "HEADS" and the number of words "TAILS." That difference will also probably be on the order of 1,000,000, and it is also very, very unlikely to be less than 1,000. – Tanner Swett May 24 '21 at 12:43
  • @TannerSwett Won't the relative number converge to zero? – Deschele Schilder May 24 '21 at 12:46
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    No. Like I said, that's a common misconception. When you flip a coin, it has no idea which way is towards 0 and which way is away from 0, so there's a 50% chance it'll go towards 0 and a 50% chance it'll go away from 0 (unless the total is already 0, in which case both directions are "away from 0"). So there's no reason why the relative number would converge to zero. The ratio between the number of heads and the number of tails will approach 1 as time goes on, but the difference will not approach 0. – Tanner Swett May 24 '21 at 12:50
  • @TannerSwett So the ratio of the differences will converge to zero? Or to one also? Won't the absolute number of the deviation from zero difference go to zero too? Or better, is there something that converges to zero? – Deschele Schilder May 24 '21 at 13:19
  • @DescheleSchilder If you flip the coin $n$ times, the expected heads / tails difference is (iirc) on the order of $\sqrt n$. This diverges. However, the difference divided by $n$ converges. – wizzwizz4 May 24 '21 at 14:10
  • I'm not sure exactly what you mean by "the ratio of the differences." But if you let $h$ be the number of heads you've flipped, and $t$ be the number of tails you've flipped, then the difference between $h/(h+t)$ and $t/(h+t)$ converges to $0$ (even though the difference between $h$ and $t$ does not converge to $0$). – Tanner Swett May 24 '21 at 15:19
  • Why are clouds so (relatively) steady, when the droplets in them can be moving at fairly high speeds? For instance cumulonimbus (thunder clouds) look slow & stately from the outside, but if you're a pilot, you've learned that they're full of destructive turbulence. – jamesqf May 24 '21 at 16:14
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    @TannerSwett Assuming a fair coin, the average difference will tend to zero. The count of H-T will cross zero infinitely often, and will just as likely be positive as negative at any given point (ignoring priors). – Trixie Wolf May 25 '21 at 22:52
  • @TrixieWolf The count of $h - t$ will cross every number infinitely often. There's nothing special about 0 except that it's the initial state, and the initial state becomes less and less relevant as the process progresses. – Tanner Swett May 26 '21 at 00:33
  • @TannerSwett That generalization is also correct, but I don't agree completely with the characterization. The starting point is still special because most of the time the tally will fall between sqrt(n) distance of that starting point regardless as to how far the sequence progresses. While that is eventually also true for any tally point you reach during the infinite sequence, convergence is going to be slower for arbitrary tally points far outside of sqrt(n) if you don't ignore all the previous flips because you'll have to wait for those flips to have a negligible contribution. – Trixie Wolf Jul 13 '21 at 16:56

3 Answers3

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The color depends on the relative angle between you, the drop, and the sun.

If you were to track an individual drop it would change color as it falls "through" the rainbow. That would be cool to see!

In more detail:

Consider a cartoon similar to the one in @John Hunter's answer. There is a viewer at left looking right. There is a sun behind the viewer. For every point to the right of the viewer we can draw lines from that point to both the viewer and the sun. So we correspond an angle to EVERY point in space.

We can ask "what are the surfaces of constant angle?" Well, imagine a line through the sun and the viewer. All points along this line will be angle zero. Larger angles will be cones coaxial with this center line. These are the surfaces of constant angle.

A rainbow works because, for certain angles, the droplet will reflect the various spectral components of the sun to the viewer.

So what is fixed in space, given the position of the viewer and sun, are the cones of constant angle. These are present whether or not there is water. A rainbow arises when there are water droplets occupying the appropriate cones.

As a droplet falls, it passes through the different cones, meaning the component of the sun that it reflects changes in time as it passes through the cones of differing angles.

Chappo Hasn't Forgotten
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Jagerber48
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    The "cool to see" part is pretty much doable. Use the garden hose in a sunny day. You can see the colorful droplets passing thru the rainbow. – fraxinus May 24 '21 at 06:59
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    "would be cool to see" — here you are ;) – Ruslan May 24 '21 at 07:05
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    Upon further thought, this answer somehow reverses the situation (though I can see what you mean!). It states that the rainbow is already there and the droplets change their color (as seen by us) to conform to the colors of the bow. Obviously, they change color but how do these changing colors constitute the rainbow in the first place? I can see, when all droplets are static in a vertical plane, how the bow appears, and upon falling the drops will "fall through". But what about the horizontal direction? Will not different parallel vertical planes give different rainbows? – Deschele Schilder May 24 '21 at 15:06
  • @DescheleSchilder which vertical planes do you mean? Flat side to the observer, differing by distance? – Ruslan May 24 '21 at 16:10
  • @Ruslan I think that what you describe is what I mean, yes. Say, there are two parallel planes with droplets in front of you. Your distance to the right part of these planes is the same as to the left part. The planes are perpendicular to the Earth. – Deschele Schilder May 24 '21 at 16:21
  • @DescheleSchilder rainbow is an angular phenomenon, so distance doesn't affect it. View direction affects the color you see, not the (Cartesian) position of the drop. If you have two parallel rectangles in front of you, the sides of one rectangle will be at a different angular separation than those of another (perspective effect). So you'll see different colors at these sides, according to their angular positions (view directions). – Ruslan May 24 '21 at 16:29
  • @DescheleSchilder I explicated the answer a lot further, it should be clear now. – Jagerber48 May 25 '21 at 00:06
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This picture was posted on this answer, Rainbows and Clouds might be handy again now

enter image description here

The drops are in continuous motion but the two shown will quickly be replaced by others in the same positions...

(although the image has been regularly used on Stack exchange: Copyright 1999 Rebecca McDowell rebeccapaton.net/rainbows/formatn.htm )

John Hunter
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  • This was in fact the picture that inspired my question! Or better, re-sparked my question. – Deschele Schilder May 23 '21 at 21:51
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    Interesting how sometimes a double rainbow can be seen, quite rare but it happens sometimes...hope this helps – John Hunter May 23 '21 at 21:54
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    There is a slight error in this image: the reflection at the inside-surface of the drop is most certainly not Total Internal Reflection, but rather just "normal" reflection. We've had extensive discussion about this, see this answer for a "corrected" version of the same image, and the question for a discussion as to why it cannot be Total Internal Reflection. – Philip May 23 '21 at 21:58
  • @JohnHunter Whenever I see a rainbow, it is almost always double. The second one has bigger visible radius, reversed color order and less brightness. The second one is created by two internal reflections. – fraxinus May 24 '21 at 06:25
  • You are lucky, they seem to be quite rare here! Yes, colour reversed for the outer one, maybe we should check how it's formed, after Philips comments... – John Hunter May 24 '21 at 07:59
  • @JohnHunter There's almost always many more than a single rainbow when it's actually raining, it's just that the others get progressively harder to see - and the contrast may make even the second one very hard to see. Essentially, for every successive reflection, you get another (fainter) rainbow. If the rainbow is caused by just a thin spray of water (e.g. waterfall or a hose), it's very unlikely you'll see more than one, though. And don't use sunglasses with polarization filters - rainbows are strongly linearly polarized :) – Luaan May 24 '21 at 10:11
  • Good point about the sunglasses and it's true there will be more than one. It just meant that the second isn't usually visible here. Probably the average light is dimmer, sunglasses are rarely worn. Although we have enough rain which is also needed! – John Hunter May 24 '21 at 12:18
  • Probably worth acknowledging the origin and copyright holder of this image! (Copyright 1999 Rebecca McDowell http://www.rebeccapaton.net/rainbows/formatn.htm) – user2705196 May 24 '21 at 14:30
  • OK, it was cut and pasted from an image without the copyright mentioned. It'll be added – John Hunter May 24 '21 at 14:35
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    @Deschele Schilder: I can see why -- the guy definitely looks nonplussed that the droplets keep moving while the rainbow stays still. – A. I. Breveleri May 24 '21 at 20:19
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The apparent colour of each particle depends on the position, but only on “2D position in view”, i.e. angular position relative to the observer. (See other answers and questions for explanation on this.) Thus it doesn't matter how chaotic they move: whereever each particle is, it will have the “correct” colour to fit in the rainbow.

Position-dependent-colour particles forming a rainbow

leftaroundabout
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