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aformalevent
2014-Apr-23, 04:12 AM
Hi everyone! It's been a very long time since I've posted anything on the forums in here but I've always loved the site (and especially the podcasts :))

So my question is, on a clear cloudless day at noon, when you look up at the blue sky, on average, how many times have the blue photons you see bounced on particles in the atmosphere before they hit your eyes? Obviously I am just looking for an estimate. I really have no idea. Assuming you're not looking directly at the sun, the photon must have bounced at least once for you to see it right? Some photons must bounce many, many more times. Can anyone give a reasonable approximation for the average with a little bit of theory to back it up? The most I can come up with is that the photons from the blue sky nearest to the sun will probably have bounced fewer times on average than those perceived as coming from the horizon. Is it 50? is it 50 million?

Here is the post I was making elsewhere which lead me to the question, if anyone is interested:
Cool thought: If the sun is pure white, why does it look yellow? It's awesome, the first astronauts above the atmosphere remarked on the brilliant white color of the sun!

Most of the suns light travels a b-line from its surface to your eyes, which is what you see when you look directly at it (not recommended). The blue light, on the other hand, bounces around randomly in the atmosphere before arriving on your retinas. So you don't perceive it as coming from the direction of the sun, but from everywhere in the sky, which is why the entire sky looks blue. So the white light of the sun is split into parts by the sky like a giant prism, with the blue light being spread out over the entire sky, and the rest, which appears yellow, coming directly from the sun.

You can see this as the sun sets. The reason the sun yellows as it moves closer to the horizon is that its light must pass through more atmosphere on rout to your eyes, so more of the blue light gets bounced around in the sky. The blue light your missing from the sun when its near the horizon is brightening the the blue skies of the people further west, who are in the middle of their day.

It's crazy to think the astronauts on the moon saw a brilliant, pure white sun shining in a perfectly black sky, but its true! No atmosphere, no blue sky, no yellow sun.

George
2014-Apr-23, 12:29 PM
There is a great book on this topic by Gotz Hoeppe, Why the Sky is Blue.

The average no. of scatters were determined to be a little more than 1 from work done at the time of Lord Rayleigh, and by him, I think. I don't know if this has been changed. Multiple scattering would actually diminish the blue sky color since proportionally more blue photons would scatter out of the atmosphere.

Regarding solar non-yellowness, more can be found from this somewhat tongue and cheek blog here (http://www.science20.com/solar_fun_heliochromologist/color_sun_revelation-30294).

George
2014-Apr-23, 12:50 PM
It's crazy to think the astronauts on the moon saw a brilliant, pure white sun shining in a perfectly black sky, but its true! No atmosphere, no blue sky, no yellow sun.[/I] The problem with their report is that the Sun, whether seen from space or from down here, is blinding in intensity. The brightness is so great it far exceeds the visual threshold limits of our color cones. If all our color cones are maxed, then I think only white will be registered even if one color is far greater than another. A neutral attenuation of 10,000x or more is needed to accurately determine directly the Sun's true color, and it needs to be done from space. [The evidence is overwhelming already, but it would be a nice way to convince others.]

Astronaut Don Pettit had planned to do a solar projection for color determination on his last ISS trip, but it did not happen. He is going back up, apparently, so we may get some dramatic additional evidence.

StupendousMan
2014-Apr-23, 12:51 PM
Light from space can bounce off atoms or dust particles as it moves through the atmosphere. Astronomers measure the fraction of blue light which makes it to the Earth's surface without bouncing to be roughly 70 percent, if the light travels vertically downward from the zenith. In other words, more than half of the blue photons don't bounce at all on their way down. Perhaps this might help you to estimate the number of bounces for sunlight.

EigenState
2014-Apr-23, 01:03 PM
Greetings,


Hi everyone! It's been a very long time since I've posted anything on the forums in here but I've always loved the site (and especially the podcasts :))

So my question is, on a clear cloudless day at noon, when you look up at the blue sky, on average, how many times have the blue photons you see bounced on particles in the atmosphere before they hit your eyes? Obviously I am just looking for an estimate. I really have no idea. Assuming you're not looking directly at the sun, the photon must have bounced at least once for you to see it right? Some photons must bounce many, many more times. Can anyone give a reasonable approximation for the average with a little bit of theory to back it up? The most I can come up with is that the photons from the blue sky nearest to the sun will probably have bounced fewer times on average than those perceived as coming from the horizon. Is it 50? is it 50 million?

I cannot give you an estimate, but I can give you some resources regarding the theory.

The relevant phenomenon is elastic light scattering. Two principal mechanisms are of interest: Rayleigh scattering (http://en.wikipedia.org/wiki/Rayleigh_scattering) and Mie scattering (http://en.wikipedia.org/wiki/Mie_scattering). See also Blue sky and Rayleigh scattering (http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html#c2).

Both mechanisms scatter visible light. Rayleigh scattering is by diatomic molecule sized species. Mie scattering is by larger, yet still small particles of dust for example. The two different mechanisms manifest different wavelength dependencies and different angular distributions. As a result, the dominant mechanism depends on the position of the Sun in the sky.

Best regards,
ES

grapes
2014-Apr-23, 01:31 PM
The average no. of scatters were determined to be a little more than 1 from work done at the time of Lord Rayleigh, and by him, I think. I don't know if this has been changed. Multiple scattering would actually diminish the blue sky color since proportionally more blue photons would scatter out of the atmosphere.
Yes, once. Otherwise, the light of the sky would not be polarized, 90 degrees from the sun.

Ken G
2014-Apr-23, 02:45 PM
Multiple scattering would actually diminish the blue sky color since proportionally more blue photons would scatter out of the atmosphere.Indeed, if there was too much scattering, the sky would actually look red! That's what is happening when you see a band of red around the horizon at sunset. Roughly speaking, the prevailing color of a given patch of sky is whatever color is expected to scatter approximately once, which also explains the degree of polarization.

George
2014-Apr-23, 04:52 PM
Yes, once. Otherwise, the light of the sky would not be polarized, 90 degrees from the sun. That is a good point. I think it was the measured brightness of the blue sky that forced them into a number slightly greater than one. I don't think, however, this conclusion was properly explained, at least not in georgeeze. My wild guess would be that if the blue photons scatter more often, then more would be lost not gained, but at the same time, more blue-green and green photons would necessarily scatter more, so since there are slightly more green photons than blue photons, brightness would increase though it would be a less saturated blue. But that seems to be what we see. Am I close?

Ken, it is interesting that we would first tend to get more of a white sky with increased scattering before we would get to a red sky. Grant made this argument some time ago that the color of the light source would be the net color result for extensive scattering, though further scattering would produce red. I suspect he is close but inverse 4th scattering, and other scattering, would not produce such a simple matching result, yet it would certainly whiten in our case.

Curiously, the sky is still blue during sunset. That too was a mystery that is now attributed to our tiny amount of ozone and is described in the Chapuis effect.

George
2014-Apr-23, 04:52 PM
Yes, once. Otherwise, the light of the sky would not be polarized, 90 degrees from the sun. That is a good point. I think it was the measured brightness of the blue sky that forced them into a number slightly greater than one. I don't think, however, this conclusion was properly explained, at least not in georgeeze. My wild guess would be that if the blue photons scatter more often, then more would be lost not gained, but at the same time, more blue-green and green photons would necessarily scatter more, so since there are slightly more green photons than blue photons, brightness would increase though it would be a less saturated blue. But that seems to be what we see. Am I close?

Ken, it is interesting that we would first tend to get more of a white sky with increased scattering before we would get to a red sky. Grant made this argument some time ago that the color of the light source would be the net color result for extensive scattering, though further scattering would produce red. I suspect he is close but inverse 4th scattering, and other scattering, would not produce such a simple matching result, yet it would certainly whiten in our case.

Curiously, the sky is still blue during sunset. That too was a mystery that is now attributed to our tiny amount of ozone and is described in the Chapuis effect.

Copernicus
2014-Apr-23, 05:25 PM
I don't know how true this is, but the following article says the sky could just as well be purple or blue, but your brain tells you it is blue.

http://www.nbcnews.com/id/8631798/ns/technology_and_science-science/t/why-skies-are-blue-instead-purple/#.U1f3MPldVqU

grapes
2014-Apr-23, 05:45 PM
Ken, it is interesting that we would first tend to get more of a white sky with increased scattering before we would get to a red sky. Grant made this argument some time ago that the color of the light source would be the net color result for extensive scattering, though further scattering would produce red. I suspect he is close but inverse 4th scattering, and other scattering, would not produce such a simple matching result, yet it would certainly whiten in our case.

Isn't that why clouds are white? More or less...


I don't know how true this is, but the following article says the sky could just as well be purple or blue, but your brain tells you it is blue.

http://www.nbcnews.com/id/8631798/ns/technology_and_science-science/t/why-skies-are-blue-instead-purple/#.U1f3MPldVqU
In this case, it's not the brain's fault, it's your eyes. :)

George, you should be all over this!

George
2014-Apr-23, 05:56 PM
I don't know how true this is, but the following article says the sky could just as well be purple or blue, but your brain tells you it is blue.

http://www.nbcnews.com/id/8631798/ns/technology_and_science-science/t/why-skies-are-blue-instead-purple/#.U1f3MPldVqU What is missing there is that the photon flux for violet from the Sun is weak relative to the rest of the spectrum (380nm photons are half those in number for 450nm). Also, the spectral sensitivity of the eye is weak in the violet end of the spectrum. Given that only about 2% of our color cones are "blue", it's amazing that we see so much of it. Many cone sensitivity graphs don't extend beyond 400nm, though violet has a range that likely reaches 420nm or so, yet blended with blue at some point.

George
2014-Apr-23, 06:06 PM
Isn't that why clouds are white? More or less... White clouds and snow fit the Mie Scattering model where all wavelengths are scattered evenly, though atmospheric scattering will affect clouds especially. Have you ever noticed how the more distant puffy clouds on a bright day are more yellowish or even pinkish? With some effort, you could probably calculate their distance based on color alone. [Please don't ask me to do so.]

I thought the purple blend to blue was odd, but the retinex thing is powerful. Since purple is red + blue, I disregarded the arguments, rightly or not.


George, you should be all over this! Ok, that'll be 5 bucks, assuming most of it is correct.

Strange
2014-Apr-23, 09:28 PM
White clouds and snow fit the Mie Scattering model where all wavelengths are scattered evenly ...

When I was on a glacier, the snow was quite an intense blue. I'm not sure why. Because there was a lot of ice as well, maybe?

Jeff Root
2014-Apr-24, 09:37 AM
And looking down from an airplane, the open spaces between
clouds were dark blue.

Also, there was a band of green between the top of the white
cloud layer and the blue sky above.

-- Jeff, in Minneapolis

George
2014-Apr-24, 01:59 PM
When I was on a glacier, the snow was quite an intense blue. I'm not sure why. Because there was a lot of ice as well, maybe?That had to be impressive. I wonder if the direct Sun was blocked and you were getting reflected blue sky. The Mie Scattering effect for the snow would reflect the blue sky color like a mirror.

Take a sheet of white paper and stand in the shade during a bright blue sky day, then watch the white paper quickly turn to a distinctive tint of blue.

Shaula
2014-Apr-24, 05:28 PM
That had to be impressive. I wonder if the direct Sun was blocked and you were getting reflected blue sky. The Mie Scattering effect for the snow would reflect the blue sky color like a mirror.
If you get very pure ice, such that light scatters volumetrically from it, then the blue colour is probably at leaast partly due to absorbance. Water absorbs more red - by nearly a factor of 100.

Jeff Root
2014-Apr-24, 06:39 PM
I think the green band I saw in the sky was the same color as
the green tint of the water in my white bathtub when I fill it
deep enough. Also the same as the green tint seen through
the glass of about a dozen stacked storm windows.

-- Jeff, in Minneapolis

NEOWatcher
2014-Apr-24, 06:45 PM
I think the green band I saw in the sky was the same color as
the [...] the same as the green tint seen through
the glass of about a dozen stacked storm windows.
I hope you're talking about the look of the shade of green and not speculating on causes. Unless there's a lot of soda-lime and iron oxide up there.

Jeff Root
2014-Apr-24, 07:06 PM
I don't know (or have forgotten if I did know) what causes
the green color in glass, so as far as I *do* know, it could
be the same cause. I'm not pushing the idea, just putting
it out as a possibility, being certain that some of you *do*
know. A nice, big article in Scientific American tells about
the whole range of causes of color, but I don't remember
it specifically mentioning window glass.

-- Jeff, in Minneapolis

George
2014-Apr-24, 07:13 PM
If you get very pure ice, such that light scatters volumetrically from it, then the blue colour is probably at leaast partly due to absorbance. Water absorbs more red - by nearly a factor of 100. That is true for ice, and the blue ice pictures from Antarctica are beautiful. Oceans are blue for the reason you state, though some have claimed otherwise.

But I don't know how snow would behave like ice. Are glacier snow flakes somehow larger or mineralized to absorb light as water does?

George
2014-Apr-24, 07:18 PM
I don't know (or have forgotten if I did know) what causes
the green color in glass, so as far as I *do* know, it could
be the same cause. I'm not pushing the idea, just putting
it out as a possibility, being certain that some of you *do*
know. A nice, big article in Scientific American tells about
the whole range of causes of color, but I don't remember
it specifically mentioning window glass.

I wonder if the bathtub green might be due to the use of a typical incandescent light bulb. They produce more green light than blue, no doubt. The filaments are at less than half the temperature of the Sun, and do have blackbody (sorta) SEDs.

NEOWatcher
2014-Apr-24, 07:19 PM
I don't know (or have forgotten if I did know) what causes
the green color in glass, so as far as I *do* know, it could
be the same cause.
I'm not sure about that either, but I know it's not from water.

EigenState
2014-Apr-24, 07:36 PM
Greetings,


I don't know (or have forgotten if I did know) what causes
the green color in glass ...

Iron impurities.

Best regards,
ES

Jeff Root
2014-Apr-24, 07:39 PM
I know it's not from water, too. But while definitely not
the same material, it *could* be the same mechanism.

OH! I just remembered something about OH ions or the
like in water being responsible for the color. Maybe.

-- Jeff, in Minneapolis

swampyankee
2014-Apr-27, 12:55 PM
As a rough estimate for how many times a blue photon bounces, one could probably calculate the mean free path vs altitude and calculate the resulting frequency of collision vs altitude, and just integrate that.

Have fun; I'm too lazy.

grapes
2014-Apr-27, 06:45 PM
As a rough estimate for how many times a blue photon bounces, one could probably calculate the mean free path vs altitude and calculate the resulting frequency of collision vs altitude, and just integrate that.

Have fun; I'm too lazy.
For solar photons, the mean free path is nearly infinite. Most photons don't get bounced, the ones we see directly from the sun. The rest of the sky pales in comparison. So the resulting frequency of collision is nearly zero.l

StupendousMan
2014-Apr-27, 08:01 PM
Post #4 in this thread provides evidence based on astronomical observations that only 70% of photons from the Sun make it through the Earth's atmosphere without scattering; that's if the Sun is directly overhead -- a small fraction make it through when the Sun is lower in the sky.

Do you have any evidence for your claim?

Jeff Root
2014-Apr-27, 08:26 PM
In post #4 you said the 70% was "the fraction of blue light".

I don't think that will help grapes get to a nearly infinite free
path, though... Maybe for red light??

-- Jeff, in Minneapolis

grapes
2014-Apr-27, 09:34 PM
Post #4 in this thread provides evidence based on astronomical observations that only 70% of photons from the Sun make it through the Earth's atmosphere without scattering; that's if the Sun is directly overhead -- a small fraction make it through when the Sun is lower in the sky.

Do you have any evidence for your claim?
My claim? The question in the OP is "how many times have the blue photons you see bounced on particles in the atmosphere before they hit your eyes?" The answer seems to be "once", right? The OP seemed to be thinking the answer was more than one, even much greater than one. If the answer (for all photons) is .3, that's nearly zero.

I guess "nearly infinite" and "nearly zero" are relative terms... :)

Hornblower
2014-Apr-27, 10:41 PM
The fact that some 30% of the incoming blue light is scattered, mostly in the last few miles before reaching the ground, suggests to me that the mean free path is at most a few tens of miles, which is a far cry from being virtually infinite.

If I am not mistaken, the scattering is approximately isotropic, as indicated by the blue haze that washes out views of the ground from a high flying aircraft or a satellite. Thus roughly half of the 30% that is scattered goes back out into space, while the rest reaches the ground to be seen as our familiar blue sky. That is in pretty good agreement with my own measurements with an exposure meter, which shows the total light from a clear sky as roughly 10% of what comes directly from the Sun. Of all the photons that reach us from the sky after the initial bounce, I would expect the majority of them to reach our eyes after no further bounces, with a minority undergoing two or more bounces. My educated guess is that the average number of bounces is well under two.

Don't forget that plenty of red, orange, yellow and green light is scattered along with the blue and violet, as shown by the pastel nature of the blue sky, and the fact that objects of all colors look largely normal in skylight from a north facing window.

Once again, don't take this as gospel, as I am by no means an expert, but I do have high confidence in my reasoning.

George
2014-Apr-28, 01:16 AM
As a rough estimate for how many times a blue photon bounces, one could probably calculate the mean free path vs altitude and calculate the resulting frequency of collision vs altitude, and just integrate that. It's not so simple. The problem with our atmosphere is that, at sea level, the distance to the air particles are about one hundredth the wavelength of visible light. This creates destructive interference so that air becomes transparent, and we should see a dark sky instead of a blue one. Einstein and Smoluchowski (1910), working with critical points, applied their work to show that the Poisson statistics apply where, for N particles, there is an N plus or minus sq. rt. N for any region. This variation allows for non-destructive scattering. They also, remarkably, derived the inverse 4th law mathematically, independent of Rayleigh's work.

Ken G
2014-Apr-28, 12:21 PM
That's interesting, it sounds to me like you are saying you can get the 1/lambda to the 4th law by looking at the variance in the number of particles within the wavelength of the light and applying refraction theory. I didn't know that. It adds to the two other ways you can also get it-- you can get it by looking at a single bound electron, and shake it fast with an electric field of given strength, and you can get it by looking at the coherence in the oscillations within an extended object of given size that is much less than the wavelength. Note these latter two ways apply to single atoms, or to relatively huge particles, so are distinct situations. Now add the variance in the refraction and you get scattering by a collection of either of those, and we have 3 completely different situations where you get the Rayleigh law, all of which are going on in the air! I'm not sure if they key here is to find the one that dominates, and then that is the "correct" explanation of the blue sky (which you are saying is the third of those, the one few people know about except for the great one), or look for the unifying principle here that, classically, they all must share the key element of Rayleigh scattering, which is that you have the oscillation of some kind of effective dipole, where the strength of the dipole does not depend on the frequency of the light. This behavior occurs when the dipole in some sense reacts quickly enough to always maintain a state of instantaneous equilibrium with the electric field to which it is exposed, regardless of frequency.

George
2014-Apr-28, 03:30 PM
That's interesting, it sounds to me like you are saying you can get the 1/lambda to the 4th law by looking at the variance in the number of particles within the wavelength of the light and applying refraction theory. Sorta, because the number of particles are taken out of the equation, not that it was implying that the particles aree not there, and treated as if the medium is continuous, so that refraction is all that is necessary to get the inverse 4th relationship. I thought you would appreciate this cool tid bit given its example of unification. :)


I didn't know that. It adds to the two other ways you can also get it-- you can get it by looking at a single bound electron, and shake it fast with an electric field of given strength, and you can get it by looking at the coherence in the oscillations within an extended object of given size that is much less than the wavelength. Note these latter two ways apply to single atoms, or to relatively huge particles, so are distinct situations. Now add the variance in the refraction and you get scattering by a collection of either of those, and we have 3 completely different situations where you get the Rayleigh law, all of which are going on in the air! I'm not sure if they key here is to find the one that dominates, and then that is the "correct" explanation of the blue sky (which you are saying is the third of those, the one few people know about except for the great one), or look for the unifying principle here that, classically, they all must share the key element of Rayleigh scattering, which is that you have the oscillation of some kind of effective dipole, where the strength of the dipole does not depend on the frequency of the light. This behavior occurs when the dipole in some sense reacts quickly enough to always maintain a state of instantaneous equilibrium with the electric field to which it is exposed, regardless of frequency. I am rushed and want to come back to this, but, at a glance, I don't understand why frequency would not be critical.

Hornblower, the distribution is not isotropic since transverse waves do not induce much at 90 deg. of propogation, perhaps a little less than half close to 90 deg. instead. This was first seen in Strutt's mechanical model before Maxwell connected the electromagnetic unification to light, thus optics. That connection was unexpected.

George
2014-Apr-28, 03:34 PM
BTW, by reversing the scattering values, the number of scattering particles (Avogadro's number) could be roughly determined. Strutt knew that the British army could see the white peaks of Mt. Everest from a distance of 160 km. He had calculated that the could see them from as far away as 230km. His value came to an order of 10^19, but other results were improved. [At the time, it was not clear that a specific value would exist for mole-grams.]

Ken G
2014-Apr-29, 12:19 PM
I am rushed and want to come back to this, but, at a glance, I don't understand why frequency would not be critical.Frequency is critical in the sense that we have scattering in proportion to frequency to the 4th power, but the way you get that is to make dipoles oscillate whose strengths are independent of frequency. The amplitude of the oscillation depends only on the intensity of electromagnetic field, not on its frequency. The frequency to the 4th power effect all comes from the fact that a faster oscillating dipole, of fixed amplitude, is a faster radiating dipole! So the defining feature of Rayleigh scattering is that how much is coherently oscillating doesn't depend on frequency, only how fast it is doing that coherent oscillation.

Hornblower
2014-Apr-29, 01:57 PM
Sorta, because the number of particles are taken out of the equation, not that it was implying that the particles aree not there, and treated as if the medium is continuous, so that refraction is all that is necessary to get the inverse 4th relationship. I thought you would appreciate this cool tid bit given its example of unification. :)

I am rushed and want to come back to this, but, at a glance, I don't understand why frequency would not be critical.

Hornblower, the distribution is not isotropic since transverse waves do not induce much at 90 deg. of propogation, perhaps a little less than half close to 90 deg. instead. This was first seen in Strutt's mechanical model before Maxwell connected the electromagnetic unification to light, thus optics. That connection was unexpected.
My bold. I said approximately, and I mean a rough approximation based on my own visual sensation of generally uniform brightness over the vast expanse of sky, and of the underlying air when looking at the ground from an airliner in the stratosphere. I know that the light is strongly polarized 90 degrees from the Sun and much less approaching 0 and 180 degrees.

George
2014-Apr-29, 11:01 PM
Frequency is critical in the sense that we have scattering in proportion to frequency to the 4th power, but the way you get that is to make dipoles oscillate whose strengths are independent of frequency. The amplitude of the oscillation depends only on the intensity of electromagnetic field, not on its frequency. Ok, this looks like a power thing, which is important to the intensity of the light we observer. Is that close?


The frequency to the 4th power effect all comes from the fact that a faster oscillating dipole, of fixed amplitude, is a faster radiating dipole! So the defining feature of Rayleigh scattering is that how much is coherently oscillating doesn't depend on frequency, only how fast it is doing that coherent oscillation. This is still unclear since speed of oscillation is frequency, so I'm missing something here.

George
2014-Apr-29, 11:03 PM
My bold. I said approximately, and I mean a rough approximation based on my own visual sensation of generally uniform brightness over the vast expanse of sky, and of the underlying air when looking at the ground from an airliner in the stratosphere. I know that the light is strongly polarized 90 degrees from the Sun and much less approaching 0 and 180 degrees.I rushed my comment and it was more curt than appropriate. My point is that that the scattering distribution, important to the topic, is diminished as it approaches 90 degrees from the Sun.

I read that the variation in brightness from noon to dusk is something like, going on memory, 70 million.

Ken G
2014-Apr-30, 03:43 PM
Ok, this looks like a power thing, which is important to the intensity of the light we observer. Is that close?Yes, it's a power thing-- the faster you shake the dipole, the more power it radiates, like frequency to the 4th if the strength of the dipole doesn't depend on frequency.


This is still unclear since speed of oscillation is frequency, so I'm missing something here.Frequency is rate of oscillation-- amplitude of oscillation is something quite different.

grapes
2014-Apr-30, 04:20 PM
I had the same reaction that George did.

The frequency to the 4th power effect all comes from the fact that a faster oscillating dipole, of fixed amplitude, is a faster radiating dipole! So the defining feature of Rayleigh scattering is that how much is coherently oscillating doesn't depend on frequency, only how fast it is doing that coherent oscillation.
I'm having trouble parsing that last sentence. :)

What is the defining feature of Rayleigh scattering? It looks like the last part of the sentence is just saying this:


Frequency is rate of oscillation-- amplitude of oscillation is something quite different.

Ken G
2014-May-01, 01:59 AM
I'm a bit unclear on what the confusion is, so I'll just explain Rayleigh scattering and it all should become clear. I'll give one way to explain it, anyway, although it sounds from the above like there are three quite different situations where it happens. Just consider a positive and negative charge connected by a spring. Let the positive charge have a lot more mass, so we can just imagine it stays in place, and the negative charge is free to "orbit" around the positive charge, at whatever distance the spring says it should (the spring force is proportional to distance). Now hit this system with a circularly polarized electric field (that it actually the simplest situation, just imagine an electric vector of fixed length that is going around in a circle). The electric vector is amplitude E and frequency f. Also imagine that f is slow enough that the spring has plenty of time to reach an equilibrium with the electric force, i.e., the spring force kr (for spring constant k and radius r) is equal to the electric force eE. That's it, that's Rayleigh scattering. The electron is just a dipole of amplitude d=er, and since we have r=eE/k that means the dipole is d=Ee^2/k. So the dipole strength is proportional to the electric amplitude E, but independent of frequency f. The rate it emits light is given by the "Larmor formula", which says the power is proportional to the square of the second time derivative of the dipole, so that's (d*f^2)^2, since the time rate of change of d is d*f. So the power scales like d^2*f^4, so that's like E^2*f^4. To get a scattering cross section, we need that the flux of energy in the light scales like E^2, so it's just the flux of energy in the light times f^4. That's Rayleigh scattering-- the oscillation amplitude is independent of frequency, but that automatically gives power like f^4, because things that oscillate faster emit more light like f^4, all else being equal.