Question about Rainbow curvature

My wife and I saw a rainbow this evening and it got me thinking.

Why does a rainbow cure so much that it looks like it touches down somewhere? Why not follow the curve of the earth and be just suspended in the air? Come to think of it, why is a rainbow curved at all.

Rick the curious.

Curiosity killed the cat you know…

Sorry

ducking behind my rock

Rainbows are actually circular, as you can see by taking a regular garden hose and spraying a fine mist on a sunny day.

As for why they’re curved, it all has to do with angles of refraction. The angle (with respect to the sun) that is needed to refract a particular color can be achieved in any direction, so each color band in a rainbow occurs at the correct angle in all directions, creating a circle.

However, because of the way the sun is positioned in the sky, ay least half of the ranbow-circle normally occurs at an angle where there are no suspended water droplets, so the rainbow stops.

More info here: http://science.howstuffworks.com/rainbow.htm

As an interesting side note, the trivia question “What is at the center of the rainbow?” is answered “your shadow.” The reason that a rainbow typically looks like an arc is because your shadow is typically very low. When you see a rainbow from an airplane (sunlight on clouds below you), it’s a circle around the shadow of the plane.

Rainbows are circular because raindrops are spheres. Also see:

Randrops are not teardrop-shaped
http://www.ems.psu.edu/~fraser/Bad/BadRain.html

No, it has to do with the fact that the angle of refraction is constant for a given color, but the direction of refraction isn’t. Imagine you’re firing a gun at the center of a clock face, and due to the nature of the gun, all the bullets leave the barrel at an angle of 10 degrees from true. Some go towards 12 o’clock, some go towards 2, but they all go exactly 10 degrees from the center. The pattern these bullets will trace out on the wall will be a circle.

With light, each wavelength (color) deflects by a different (but specific) amount when it passes through a raindrop, so each color’s circle is a different radius.

Raindrops could be shaped like little triangular prisms and you’d still end up with circular rainbows.

Originally posted by ntucker

Nope.

Triangular raindrops would not form circular rainbows. I don’t think they would form visible spectra coherently to any given point.

Rainbows are produced by the refractive and internal reflective properties of spherical drops of water. Ever see a snowbow?

Tris

You’re right, I think that was a bit hasty. I was thinking that, given a cloud of randomly-oriented prisms, the ones which were oriented properly to bend the light by the right amount would make up the rainbow.

I still contend that the shape of the rainbow is not governed by the shape of the raindrop. The shape of the raindrop may make the phenomenon possible, but the shape of the rainbow itself is governed by the simple fact that light bending from its target by a fixed angle in a random direction will create a circle. If not with prisms, this should be possible with other shapes besides spheres (e.g. ellipsoids), shouldn’t it?

Nope.

The rainbow does not depend only on the intersection of light incident with the raindrops from a point source. It depends on the alignment of the reflected and refracted individual colors returning back toward the observer. In order for that to create the same color sorted spectrum for one observer for every raindrop, all of the raindrops must be symmetrical with respect to the point source, and the observer. The only shape that fulfills that definition is the sphere.

Tris

"It depends on the alignment of the reflected and refracted individual colors returning back toward the observer. "

I’d like to ammend some tweaks to support ntucker.

I think ntucker is coming from the perspective that if you had unspherical drops (e.g. ice crystals), the * random distribution* of the particles would still hold an overall statistical consistancy that would produce a rainbow-type phenomenon. However, the radius of the bow, as well as its refractive spectrum distribution, would not be the same as a water-droplet bow.

So this statement “In order for that to create the same color sorted spectrum for one observer…” isn’t necessarily true. As long as the macro-scale statistical distribution of the particles (even if unspherical) is the same for the observers, then they will see the same phenomenon.

Nope.

Well, they will see some sort of phenomenon, but it won’t be a rainbow.

You see the light from each raindrop only if the sun’s light is refracted and reflected from directly behind you, to the drop, and then back to you. The sorting of colors comes from the angle from the sun, to each drop, and then to you. Each region of angular refraction and reflection causes you to see a different color, within a narrow range of angles, and the fact that all the drops are spheres makes the region for each color the same, with respect to each observer.

Change the shape, and the regions will not be contiguous for each every drop, and each and every observer, because the sphere is the only three dimensional shape which is symmetrical for all angles of incidence and return.

Tris

Tell me what part of this explanation is wrong.

Each raindrop has light passing through it and refracting into a spectrum. The color of light passing through a specific raindrop and into your eye depends on the angle at which the light must be bent in order to take the path to your eye. So a particular spherical raindrop will, to your fixed eye, appear red if it’s one position, and blue if it’s in another, etc. This exact same phenomenon will occur with an ellipsoid raindrop, but the shape of the spectrum being emitted will be distorted. However, due to the random distribution of orientations of the ellipsoid raindrops, the overall effect should be the same.

ntucker,

Because raindrops are spherical, all the drops in one band happen to send green light toward your eyes, and all the ones in another band send red light. They will send other colors of light to other observers. The importance of spherical raindrops is that a sphere is symmetrical along every possible axis. That means that all the drops in any one band (that locus defined by the location of the sun, and the observer) will be the same angle from the center of the cone defined by the refraction characteristic of water.

The thing you see, which you call a rainbow is, in fact a very large number of raindrops. When the sun shines on them, the sun light enters the drop, is refracted, reflects off the inner surface of each drop, and then is refracted again as it leaves the other side of the drop, now headed back toward you. If the drops are not spherical, then the attitude of each raindrop will be random. If the attitude of the raindrop is not uniform, the light entering, and leaving will be refracted in a different direction.

Non spherical drops would not have a uniform optical path with respect to an outside observer. Although each would refract and reflect light in a particular direction, that direction would not be uniform among different drops, since the drops were not symmetric along a single axis. Spheres are the only shape which can be uniform among millions of individual drops in terms of angle of entry, reflection, and exit from any direction. No other geometry has the characteristics needed.

I don’t know how to be clearer, right now. The shape of raindrops is essential to the phenomenon. The spectra created by prisms, lenses, and such are called rainbows, but multiple prisms or lenses would have to be arranged very carefully to produce the same effect. Spheres cannot be arranged in any way that is not symmetric with respect to two directions.

Tris

Fair enough. It still sounds to me like the phenomenon is possible with other shapes besides spheres, but that spheres make conditions perfect and therefore increase the likelihood. But I’ll concede that I could be wrong.

A question for you: a given sphere will not bend all of the rays of a particular wavelength by the same angle – it depends on where the ray hits the surface of the sphere (i.e. if it hits dead center, it will go straight through, if it hits very close to the center it will go almost straight through, etc). So is it not actually untrue that “all the drops in one band send green light toward your eyes”? My understanding is that some rays of blue light hit those green-band drops at the right spot to refract the blue light toward your eye too, but that those are insignificant compared to the green rays coming from that band, and so you see that section as green.

In other words, my understanding is that all colors of light are coming from that band, but the green light is overwhelmingly more significant than any other color, so the band looks green.

Is that not true?

Just to throw a monkey wrench into this discussion, raindrops aren’t spherical, either. They’re a lot closer to spheres than they are to the traditional teardrop shape, but they’re flattened a bit by air resistance. It’s occured to me that, in principle, one should be able to deduce the shape of the drops from careful measurements of the bow, but I’ve never been able to measure a rainbow precisely enough to pull it off.

It is not just refraction. It is refraction, reflection, and refraction. The small portion of light that is refracted at an angle that allows it to reflect off the interior surface of the raindrop will be overwhelmingly of the same frequency, and be refracted again off the same small portion of the surface at its characteristic angle. All the light directed at any particular point along the axis where the rainbow is visible will be of nearly identical frequency, and light of other frequencies will be directed at other angles, or pass through the raindrop in random directions.

Minor deviations from spherical are what make rainbows look fuzzy. Some raindrops in any given region will be deformed enough not to send a prism back. I am not really sure what the proportions are.

I have to admit, I am surprised no one else has posted to give a more precise description, since I am sure my knowledge of optics is less comprehensive than that of some of our other members.

Tris

Plus the phenomenon seen from aircraft above undercast is known as the glory. It has a slightly smaller angular diameter than a true rainbow.

As an historical aside it’s how fighter pilots know where to fly to come out of the sun. You ‘fly’ the glory over the opponent’s aircraft and then go in for the kill…

I refer you to a wonderful site about particles in the air-
http://www.sundog.clara.co.uk/droplets/light.htm
also sundogs, glorys, rayleigh scattering- it is excellent (and I am not connected with it btw)

I’ve been thinking about whether you could have a rainbow with any other shape, trying to come up with a case which is simple enough to analyze. I believe you could get a recognizable circular rainbow from long, cylindrical raindrops falling with random orientation. It’s hard to describe without pictures and equations, so the explanation below is kind of confusing.

With a cylindrical raindrop, ignoring the effect of the ends (which is what I mean by “long” above), the refraction-reflection-refraction path (RRR path) will be specular with respect to the axis of the raindrop, and behave like a spherical raindrop in the plane perpendicular to the raindrop axis. This means the raindrops contributing significantly to an observer will all lie in a plane (the plane is the one a flat mirror reflecting to the observer would be in). This makes the case simple enough to analyze, since the orientation of the raindrop can be described with a single angle, phi.

Consider a global coordinate system, with the Sun at the -Z axis, and with theta[sub]R[/sub] the angle from the +Z axis to the raindrop. Take Phi = 0 to be the orientation when the raindrop is perpendicular to the plane including the Sun, the drop, and the observer. In this case, the RRR path in the cylindrical drop will be similar to one in a spherical drop. If instead the drop is parallel to the Sun, drop, observer plane, the RRR path would be analagous to the path of a spherical drop at the +Z axis, so no visible light is sent to the observer from the RRR path. In general, the RRR path corresponds to that of a spherical drop, but as Phi varies from 0 to 90 degrees, the angle theta[sub]S[/sub] of the corresponding spherical drop varies from theta[sub]R[/sub] to 0.

Assume the drop is at an angle theta[sub]R[/sub] where Red light following the RRR path is going to the observer when Phi = 0. If we allow the orientation angle Phi to change, the color of the light going to the observer will vary from red through all the colors to violet, then no visible light. The change in the color of light will be second order near Phi = 0, so there will be more red following the RRR path, then less of the colors that are farther from red.

If the drop is at a position where a raindrop with Phi = 0 is reflecting yellow light, as Phi changes from zero, light of green, blue, and violet will go to the observer (but not as much as yellow), and there will be no orange or red light. At an angle where violet goes to the observer when Phi = 0, no other colors will follow the RRR path to get there.

So in this case, there will be color variation where you’d see a rainbow from spherical drops, probably the same colors but washed out.

Sorry to drag up an old thread, but this EPOD picture of an Earthbound Rainbow is just too cool to go unseen.