Color of sky on human friendly planets
Jabir Amirali
I was watching an old Star Trek episode with Kirk and the gang walking around
on a planet with a red atmosphere, and I got to thinking...
Why is our sky blue? I mean I know that the atmosphere scatters blue light.
But why blue? Is it inevitable that any nitrogen/oxygen atmosphere will
always be blue?
Or can we breathe an atmosphere with oxygen + some other gas besides nitrogen?
How significant is the light of the central star(s)?
Can we definitely rule out any colors, or is any color possible given the
right circumstances?
Thanks.
Paul Schlyter
In article <cygnusEL...@netcom.com>,
Jabir Amirali <cyg...@netcom.com> wrote:
> Why is our sky blue? I mean I know that the atmosphere scatters blue light.
> But why blue? Is it inevitable that any nitrogen/oxygen atmosphere will
> always be blue?
> Or can we breathe an atmosphere with oxygen + some other gas besides nitrogen?
you replaced the oxygen too with some other (transparent) gas, the
sky would still appear blue.
Our sky is blue because the air molecules are much smaller than the
wavelength of light. When light encounters particles much smaller
than its wavelength, the scattered intensity is inversely
proportional to the 4'th power of the wavelength. This is called
"Rayleigh scattering", and it means that half the vaewlength is
scattered with 2**4 = 16 times more intensity. That's why the sky
appears blue: the blue light is scattered some 16 times more strongly
than the red light. And this is also why the setting Sun appears
red: the blue light has been scattered away from the direct sunlight.
If you want another color of the sky, you need bigger particles in the
air. Replacing oxygen or nitrogen with some other gases won't make much
difference, because these molecules too will be much smaller than the
wavelength of light. You need something bigger than molecules in the
air. There is such a thing -- it's called dust.
Dust particles can be many times larger than air molecules but still
small enough to not fall out to the ground. If the dust particles
are much larger than the wavelength of light, the scattered light
will be neutral in color (i.e. white, or gray) - this happens in
clouds, which consist of water droplets. Finally, if the dust
particles are of approximately the same size as the wavelength of
light, the situation gets complex, and all sorts of interesting
scattering phenomena may happen. This happens here on Earth from
time to time, particularly in desert areas, where the sky may appear
white, brown, and sometimes even in some other color. And that's the
way it is on Mars, where the sky is pinkish, as we know from images
from Mars landers.
If the atmosphere contains lots of dust, the direct light from the
Sun or Moon may occasionally get some quite unusual color. Sometimes,
green and blue moons have been reported. These phenomena are quite
rare though -- they happen only "once in a blue moon"....
> How significant is the light of the central star(s)?
dominating illumination and perceives it as "white", within a quite
wide range of possible colors. During daytime, we perceive the light
from the Sun (6000 K) as white, and at night we perceive the light
from our incandescent lamps (2800 K, like a late, cool M star) as
white. Only if we put these two lights side-by-side, at comparable
intensities, will we perceive a clear color difference.
If the Sun was a hot star (say of spectral type B), it's likely we
still would perceive its light as "white".
> Can we definitely rule out any colors, or is any color possible given
> the right circumstances?
color likely to appear from scattering in dust or gas is the case we
have here on Earth: rayleigh scattering from gas molecules, giving
the sky a strong blue color. You won't find a sky with a similarly
strongly saturated red, yellow or green in any sky one can think of.
--
----------------------------------------------------------------
Paul Schlyter, Swedish Amateur Astronomer's Society (SAAF)
Grev Turegatan 40, S-114 38 Stockholm, SWEDEN
e-mail: pau...@saaf.se paul.s...@ausys.se pa...@inorbit.com
WWW: http://spitfire.ausys.se/psr -- updated daily!
Uncle Al
Jabir Amirali wrote:
>
> I was watching an old Star Trek episode with Kirk and the gang walking around
> on a planet with a red atmosphere, and I got to thinking...
>
> But why blue? Is it inevitable that any nitrogen/oxygen atmosphere will
> always be blue?
> Or can we breathe an atmosphere with oxygen + some other gas besides nitrogen?
>
> right circumstances?
"The Physics and Chemsitry of Color" Kurt Nassau
The sky is colored blue by light scattering, Rayleigh (particles smaller
than lambda) and Mie (particles larger than lambda). The Blue Ridge
Mountains are blue because of scattering from photochemical smog
droplets produced from isoprene generously emitted by heat-stressed
conifers.
If the central sun's light contains blue, a planet's atmosphere will be
blue. The original pictures of a red Martian sky were part artifact
(poor color calibration) and part the low angle of the sun in the sky
(as with a Terran sunset).
If the planet orbits a red star whose light is deficient in blue, then
you cannot scatter what you do not have. Sky color will still be offset
to shorter wavelengths vs the mix of incoming light.
--
Uncle Al Schwartz
Uncl...@ix.netcom.com ("zero" before @)
http://pw2.netcom.com/~uncleal0/uncleal.htm
http://www.ultra.net.au/~wisby/uncleal.htm
http://www.guyy.demon.co.uk/uncleal/uncleal.htm
(Toxic URLs! Unsafe for children, Democrats, and most mammals)
"Quis custodiet ipsos custodes?" The Net!
Rainmaker
Paul,
That was an excellent explaination to an age old question. I learned something. Can
you explain the similarities and differences in the properties of light when making
the ocean that deep deep (saturated) blue color? Is it exacly like the sky, but
darker because there is less light? Or are water molecules bigger?
Gregory Loren Hansen
In article <66rgde$a...@electra.saaf.se>,
Paul Schlyter <pau...@electra.saaf.se> wrote:
>In article <cygnusEL...@netcom.com>,
>Jabir Amirali <cyg...@netcom.com> wrote:
>
>> But why blue? Is it inevitable that any nitrogen/oxygen atmosphere will
>> always be blue?
>> Or can we breathe an atmosphere with oxygen + some other gas besides nitrogen?
>
>you replaced the oxygen too with some other (transparent) gas, the
>sky would still appear blue.
>
>Our sky is blue because the air molecules are much smaller than the
>wavelength of light. When light encounters particles much smaller
>than its wavelength, the scattered intensity is inversely
>proportional to the 4'th power of the wavelength. This is called
>"Rayleigh scattering", and it means that half the vaewlength is
>scattered with 2**4 = 16 times more intensity. That's why the sky
>appears blue: the blue light is scattered some 16 times more strongly
>than the red light. And this is also why the setting Sun appears
>red: the blue light has been scattered away from the direct sunlight.
Light is scattered from density fluctuations in the air, which causes
fluctuations in the dialectric constant. A single molecule in a vacuum
would be a fluctuation, but in a volume of the diameter of a wavelength of
visible light there's tens of thousands of molecules. It's those
fluctuations in the dialectric constant that the light actually "sees"
(i.e. changes in the index of refraction). The peak, for light from the
sun, is violet light. But our eyes respond differently to different
wavelengths; the color we see isn't directly related to the energy in each
wavelength that's hitting our eye.
The shorter wavelengths (more blue) will always scatter more strongly, I
think. But if you can make the fluctuations bigger, red will be scattered
more strongly. I'm not sure how to do that, I have an experimental value
for the average fluctuations. You could also change the color of the sun.
Or become partially color-blind. Or fill the skies with various
chemicals-- Venus has sulfuric acid, I think that makes the sky look
pretty red.
>If you want another color of the sky, you need bigger particles in the
>air. Replacing oxygen or nitrogen with some other gases won't make much
>difference, because these molecules too will be much smaller than the
>wavelength of light. You need something bigger than molecules in the
>air. There is such a thing -- it's called dust.
That, too.
--
"And don't skimp on the mayonnaise!"
Michael Dworetsky
Jabir Amirali wrote:
>
> I was watching an old Star Trek episode with Kirk and the gang walking around
> on a planet with a red atmosphere, and I got to thinking...
>
> But why blue? Is it inevitable that any nitrogen/oxygen atmosphere will
> always be blue?
Yes. Pretty much any diatomic gas atmosphere should give this result.
> Or can we breathe an atmosphere with oxygen + some other gas besides nitrogen?
Sure, you could breathe a Helium-Oxygen mixture, though this is unlikely
to occur in natural situations. Seriously, I think any breathable
atmosphere on a terrestrial planet is likely to be mainly N and O for
reasons having to do with atmospheric chemistry and life. One can
conceive situations with methane-ammonia atmospheres on cold worlds, but
I think these would also have blue skies if they are not perpetually
cloudy.
If you have photochemical smog, you can get whisky-brown skies.
>
> How significant is the light of the central star(s)?
Probably not very important. Even a 'red' dwarf star is a lot bluer
than an electric incandescent light or a security floodlight. There is
plenty of blue even in the spectra of these stars.
> Can we definitely rule out any colors, or is any color possible given the
> right circumstances?
Atmospheric contaminants like dust can be very important. An example of
this is the pink sky on Mars due to dust. If you can find a valid
reason to expect green dust, then. . .
--
Mike Dworetsky, Department of Physics
& Astronomy, University College London
Gower Street, London WC1E 6BT UK
email: m...@star.ucl.ac.uk
Michael Richmann
cyg...@netcom.com (Jabir Amirali) wrote:
>I was watching an old Star Trek episode with Kirk and the gang walking around
>on a planet with a red atmosphere, and I got to thinking...
>
>Why is our sky blue? I mean I know that the atmosphere scatters blue light.
>But why blue? Is it inevitable that any nitrogen/oxygen atmosphere will
>always be blue?
>
>right circumstances?
The coloration is inevitable, regardless of gas composition, due to the
way particles scatter light. Statistical mechanics addresses this
concern.
--
Mike
My opinions, not Argonne's...
Totto
Michael Dworetsky wrote:
> > Or can we breathe an atmosphere with oxygen + some other gas besides nitrogen?
>
> to occur in natural situations. Seriously, I think any breathable
> atmosphere on a terrestrial planet is likely to be mainly N and O for
> reasons having to do with atmospheric chemistry and life. One can
> conceive situations with methane-ammonia atmospheres on cold worlds, but
> I think these would also have blue skies if they are not perpetually
> cloudy.
>
> If you have photochemical smog, you can get whisky-brown skies.
it to O3), a nitrogen/oxygen mixture (inconvenient when pressure (?)
turns it into N2O), and various helium/oxygen argon/oxygen mixtures as
far as I know.
The important thing is having oxygen + more or less non-toxic gases in
the athmosphere, that preferably don't bond to form something nasty too
often.
Am I right? :)
Totto
Boaz Ben-David
Totto wrote:
> Umm... divers breathe pure oxygen (inconvenient when pressure (?) turns
> it to O3),
Up to 2 atm partial pressure of O2 at rest or 1.6 at labor.
Hence the limit is 6 m with pure Oxygen and 100 m with air. Otherwise one gets "Oxygen
poisening".
The exposure to pure oxygen is limited in time and even at hospitals you don't get 100% O2 for
too long.
>a nitrogen/oxygen mixture (inconvenient when pressure (?)
> turns it into N2O),
At about 30M = 4Atm =3.2 partial pressure of N2, one starts to get "nitrogen narcosis".
It is very individual and some people may dive at 10 Atm, with air, without any narcosis.
>and various helium/oxygen argon/oxygen mixtures as
> far as I know.
If I recall correctly, [some 25 years elapsed] mixtures of up to 7 gases where used, including
Hydrogen.
Paul Schlyter
In article <34917F2B...@ix.netcom.com>,
Rainmaker <er...@ix.netcom.com> wrote:
>> You can rule out most highly saturated colors. The most saturated
>> color likely to appear from scattering in dust or gas is the case we
>> have here on Earth: rayleigh scattering from gas molecules, giving
>> the sky a strong blue color. You won't find a sky with a similarly
>> strongly saturated red, yellow or green in any sky one can think of.
>
> That was an excellent explaination to an age old question. I learned
> something. Can you explain the similarities and differences in the
> properties of light when making the ocean that deep deep (saturated)
> blue color? Is it exacly like the sky, but darker because there is
> less light? Or are water molecules bigger?
surface depends on many factors, such as the depth of the water, the
amount of mud (the "water equivalent" of dust), algae, etc floating
around in the water, and the color of the sea bottom (if the water is
shallow enough for this to matter here). Of course the color of the
sky matters too, as well as the angle of the incident and reflected
rays, the azimuth difference to the Sun, etc.
I cannot give a fairly simple answer to this one. Instead I'd like
to point you M. Minnaert's excellent book "The nature of light and
color in the open air", which also deals with the color of water
surfaces. This book is a little bot old, you may have to settle for
his revised "Light and Color in the outdoors", which is supposed to
be excellent too, although I haven't seen it myself.
Of course, clear water tend to give the water surface a stronger
color (deep blue, lighter blue, or green-blue, depending on circumstances)
than muddier water.
Paul Schlyter
In article <66s87p$1v5$1...@dismay.ucs.indiana.edu>,
Gregory Loren Hansen <glha...@copper.ucs.indiana.edu> wrote:
> In article <66rgde$a...@electra.saaf.se>,
> Paul Schlyter <pau...@electra.saaf.se> wrote:
>>In article <cygnusEL...@netcom.com>,
>>Jabir Amirali <cyg...@netcom.com> wrote:
>>
>>> But why blue? Is it inevitable that any nitrogen/oxygen atmosphere will
>>> always be blue?
>>
>>you replaced the oxygen too with some other (transparent) gas, the
>>sky would still appear blue.
>>
>>Our sky is blue because the air molecules are much smaller than the
>>wavelength of light. When light encounters particles much smaller
>>than its wavelength, the scattered intensity is inversely
>>proportional to the 4'th power of the wavelength. This is called
>>"Rayleigh scattering", and it means that half the vaewlength is
>>scattered with 2**4 = 16 times more intensity. That's why the sky
>>appears blue: the blue light is scattered some 16 times more strongly
>>than the red light. And this is also why the setting Sun appears
>>red: the blue light has been scattered away from the direct sunlight.
>
> Light is scattered from density fluctuations in the air, which causes
> fluctuations in the dialectric constant. A single molecule in a vacuum
> would be a fluctuation, but in a volume of the diameter of a wavelength of
> visible light there's tens of thousands of molecules. It's those
> fluctuations in the dialectric constant that the light actually "sees"
> (i.e. changes in the index of refraction). The peak, for light from the
> sun, is violet light. But our eyes respond differently to different
> wavelengths; the color we see isn't directly related to the energy in each
> wavelength that's hitting our eye.
different mechanisms. Refraction is quite predictable, and follows
the laws of geometrical optics (this is why lenses and mirrors work),
while scattering is stocastic (you cannot predict in which direction
one single photon will scatter, only the statistical distribution
of the scattered photons).
Refraction will yield its own set of phenomena, such as the apparent
flattening of the sun/moon when they are very near the horizon, the
apparent "raising" of the sun/moon at rise/set such that we still may
see them even though they "really" are below the horizon.
Scintillation ("twinkling stars") and unsteady air is also caused by
refraction. More interesting refraction effect are the mirages, in
particular the "Fata morgana". If water droplets or ice crystals
are present in the air, a whole new set of refraction-related
phenomena appear: rainbows and haloes.
A quite rare refraction effect is the "Novaya Zemlya" effect, named
so because it was first observed at Novaya Zemlya (this is Russian
for "new land"): the Sun could then be glimpsed, faintly and of
course very distorted, when it actually was some 5-6 degrees below
the horizon. Novaya Zemlya resides above the arctic circle, which
means it has midnight sun in the summer and "noon night" in the
winter. The people there know quite well at what date in January or
February the sun reappears above the horizon -- therefore a temporary
premature "return of the sun" will be quite noticeable there.
The blue color of the (clear) sky is due to scattering, not to
refraction. This is why the blue color os so smoothly distributed,
as opposed to atmospheric-optical phenomena due to refraction and
reflection: raimbows, haloes, mirages, ....
Phil Gibbs
>Why is our sky blue? I mean I know that the atmosphere scatters blue light.
>But why blue? Is it inevitable that any nitrogen/oxygen atmosphere will
>always be blue?
>Or can we breathe an atmosphere with oxygen + some other gas besides nitrogen?
You will find answers to most of your questions in the Physics
FAQ at http://www.weburbia.com/physics/blue_sky.html
Phil Gibbs
http://www.weburbia.com/ http://www.weburbia.demon.co.uk/
"The difference between a flower and a weed is a judgement."
Eric Lucas
Paul Schlyter wrote:
> In article <34917F2B...@ix.netcom.com>,
> Rainmaker <er...@ix.netcom.com> wrote:
>
> > That was an excellent explaination to an age old question. I learned
> > something. Can you explain the similarities and differences in the
> > properties of light when making the ocean that deep deep (saturated)
> > blue color? Is it exacly like the sky, but darker because there is
> > less light? Or are water molecules bigger?
>
> You're entering a quite complex subject here. The color of a water
> surface depends on many factors, such as the depth of the water, the
> amount of mud (the "water equivalent" of dust), algae, etc floating
> around in the water, and the color of the sea bottom (if the water is
> shallow enough for this to matter here). Of course the color of the
> sky matters too, as well as the angle of the incident and reflected
> rays, the azimuth difference to the Sun, etc.
>
> ...
> Of course, clear water tend to give the water surface a stronger
> color (deep blue, lighter blue, or green-blue, depending on circumstances)
> than muddier water.
These are all factors, but there is a simpler, more direct answer. Water is
inherently blue. It's apparently due to a high-energy O-H vibrational
absorption with a tail in the red end of the visible. Check out Scientific
American some time in the last 7 years for a good article describing all the
hoops the author jumped through to rule out other sources of the color
(including a 2-m IR cell full of something like $10,000 of D2O to see the
isotope shift.)
Does anyone have the exact cite for that article? I remember reading it but
not exactly when, and I've never gone through all the back issues to find it.
Eric Lucas
Gregory Loren Hansen
In article <66tlpq$d...@electra.saaf.se>,
Paul Schlyter <pau...@electra.saaf.se> wrote:
>In article <66s87p$1v5$1...@dismay.ucs.indiana.edu>,
>Gregory Loren Hansen <glha...@copper.ucs.indiana.edu> wrote:
>>>> Why is our sky blue? I mean I know that the atmosphere scatters blue light.
>> Light is scattered from density fluctuations in the air, which causes
>> fluctuations in the dialectric constant. A single molecule in a vacuum
>> would be a fluctuation, but in a volume of the diameter of a wavelength of
>> visible light there's tens of thousands of molecules. It's those
>> fluctuations in the dialectric constant that the light actually "sees"
>> (i.e. changes in the index of refraction). The peak, for light from the
>> sun, is violet light. But our eyes respond differently to different
>> wavelengths; the color we see isn't directly related to the energy in each
>> wavelength that's hitting our eye.
>
>This is plain wrong! Scattering and refraction of light are two quite
>different mechanisms. Refraction is quite predictable, and follows
>the laws of geometrical optics (this is why lenses and mirrors work),
>while scattering is stocastic (you cannot predict in which direction
>one single photon will scatter, only the statistical distribution
>of the scattered photons).
>
>Refraction will yield its own set of phenomena, such as the apparent
>flattening of the sun/moon when they are very near the horizon, the
>apparent "raising" of the sun/moon at rise/set such that we still may
>see them even though they "really" are below the horizon.
You're thinking of refraction from a uniform medium, like from a prism.
I'm talking about scattering. And the only thing that scatters light is a
change in the index of refraction; that's *the* optical property, the only
thing that does anything at all with light.
Except I'm saying there are too many molecules in a wavelength of air to
treat it as scattering from individual molecules, it would transmit as if
through glass. The scattering is from fluctuations in the air.
I.e. put bubbles in your ice.
>The blue color of the (clear) sky is due to scattering, not to
>refraction.
I said that. But you do you think the light scatters? You say molecules,
I say density fluctuations. We get the same frequency dependence, anyway.
>This is why the blue color os so smoothly distributed,
>as opposed to atmospheric-optical phenomena due to refraction and
>reflection: raimbows, haloes, mirages, ....
I thought rainbows were caused by scattering from water droplets, similar
to the effect of a diffraction grating.
Tandem Control Room
Gregory Loren Hansen wrote:
>
> In article <66tlpq$d...@electra.saaf.se>,
> Paul Schlyter <pau...@electra.saaf.se> wrote:
> >In article <66s87p$1v5$1...@dismay.ucs.indiana.edu>,
> >Gregory Loren Hansen <glha...@copper.ucs.indiana.edu> wrote:
>
> >>>> Why is our sky blue? I mean I know that the atmosphere scatters blue light.
>
> >> Light is scattered from density fluctuations in the air, which causes
> >> fluctuations in the dialectric constant. A single molecule in a vacuum
HUH? Pardon this engineers ignorance, but the dielectric constant of air
is almost exactly 1. Any reduced pressure must be also 1. Values as high
as 1.001 are possible, but what does that hace to do with light??
The dieletric constant of a material is:
"the ratio of the capacity of a condenser with that substance as
dielectric to the capacity of the same condeser with a vacuum for a
dielectric"
It is directly related to permittivity, or how easily an electric field
propogates thtough it, as compared with a vacuum.
Since there are no large electric fields present, I can't see how these
terms belong in this discusion.
I could be wrong.
I was once.
I think!
George
go...@bnl.gov
Jim Carr
Paul Schlyter <pau...@electra.saaf.se> wrote:
}
} Our sky is blue because the air molecules are much smaller than the
} wavelength of light. When light encounters particles much smaller
} than its wavelength, the scattered intensity is inversely
} proportional to the 4'th power of the wavelength. This is called
} "Rayleigh scattering", and it means that half the vaewlength is
} scattered with 2**4 = 16 times more intensity. That's why the sky
} appears blue: the blue light is scattered some 16 times more strongly
} than the red light. And this is also why the setting Sun appears
} red: the blue light has been scattered away from the direct sunlight.
glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
>
>Light is scattered from density fluctuations in the air, which causes
>fluctuations in the dialectric constant.
The original answer is correct. Rayleigh scattering.
>But our eyes respond differently to different
>wavelengths; the color we see isn't directly related to the energy in each
>wavelength that's hitting our eye.
This is true. And the fact that we associate the english word "blue"
with a wide variety of colors seen depending on other things in the
atmosphere. Larger particles, like dust, produce Mie scattering.
--
James A. Carr <j...@scri.fsu.edu> | Commercial e-mail is _NOT_
http://www.scri.fsu.edu/~jac/ | desired to this or any address
Supercomputer Computations Res. Inst. | that resolves to my account
Florida State, Tallahassee FL 32306 | for any reason at any time.
Jim Carr
>change in the index of refraction; that's *the* optical property, the only
>thing that does anything at all with light.
Not so. Look up Rayleigh scattering.
Dipoles will scatter electromagnetic radiation.
Charles Spalton
Jim Carr wrote in message <66vnb3$gro$1...@news.fsu.edu>...
>glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
>>
>>I'm talking about scattering. And the only thing that scatters light is a
>>change in the index of refraction; that's *the* optical property, the only
>>thing that does anything at all with light.
>
> Not so. Look up Rayleigh scattering.
>
> Dipoles will scatter electromagnetic radiation.
I agree; also, scattering effects can be caused by diffraction, which is I
believe the cause of the Rayleigh effect.
Charles.
Gregory Loren Hansen
In article <66vnb3$gro$1...@news.fsu.edu>,
Jim Carr <j...@ibms48.scri.fsu.edu> wrote:
>glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
>>
>>I'm talking about scattering. And the only thing that scatters light is a
>>change in the index of refraction; that's *the* optical property, the only
>>thing that does anything at all with light.
>
> Not so. Look up Rayleigh scattering.
>
> Dipoles will scatter electromagnetic radiation.
Let me quote from Panofsky and Phillips, _Classical Electricity and
Magnetism_, 2nd ed., page 415-16.
=========================================================================
"In connection with this last point, there arises a question about
random scatterers. Consider a cubical volume whose edge is half a
wavelength, and which contains many atoms or molecules. Even though the
electrons may be distributed randomly within the cube, its radiation will
be on average out of phase with that from a similar neighboring volume in
a line of observation at right angles to the beam. The net observed
scattering from many such volumes should then be zero from any angle,
except, of course, the straightforward direction. Now these are the
conditions on atmospheric scattering of visible light-- why, then, is
Rayleigh's result in accord with observation?
"The answer to this question has been given by Einstein and
Smolukowski on the basis of a study of density fluctuations in the
atmosphere. The number of scatterers per 'cell,' or pur cube of dimension
[lambda]/2, is constant only on the average, and the fluctuations in this
number are responsible for scattering.
[some math and theory omitted]
If the fluctuations are small the 'dispersion' (the square of what is
called in probability theory the root-mean-square deviation) is given by
______
(dN)^2 = N
[the "d" is a little delta, with subscript j attached to both Ns]
Therefore the total scattering is simply proportional to sum N_j=N, the
total number of scattering centers, in agreement with Rayleigh's
assumption. For denser media, the evaluation of the average of Eq.
(22-59) is more complicated, and near a chemical phase change these
fluctuations give rise to what is called critical opalescence."
========================================================================
The main point, I think, is that Rayleigh gets the correct result not
because his assumptions were right, but because of an artifact of the
statistics.
Gregory Loren Hansen
In article <3492F6...@acnsun10.rhic.BNL.GOV>,
Tandem Control Room <tvd...@acnsun10.rhic.BNL.GOV> wrote:
>Gregory Loren Hansen wrote:
>>
>> In article <66tlpq$d...@electra.saaf.se>,
>> Paul Schlyter <pau...@electra.saaf.se> wrote:
>> >In article <66s87p$1v5$1...@dismay.ucs.indiana.edu>,
>> >Gregory Loren Hansen <glha...@copper.ucs.indiana.edu> wrote:
>>
>> >>>> Why is our sky blue? I mean I know that the atmosphere scatters blue light.
>>
>--------------BIG SNIP-----------
>
>HUH? Pardon this engineers ignorance, but the dielectric constant of air
>is almost exactly 1. Any reduced pressure must be also 1. Values as high
>as 1.001 are possible, but what does that hace to do with light??
The scattering is very small. That's why you see it when you look up into
the sky, but there's no noticeable redenning of your living room lamp when
you move across the room.
Jim Carr
Jim Carr <j...@ibms48.scri.fsu.edu> wrote:
|
| glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
| >I'm talking about scattering. And the only thing that scatters light is a
| >change in the index of refraction; that's *the* optical property, the only
| >thing that does anything at all with light.
|
| Not so. Look up Rayleigh scattering.
|
| Dipoles will scatter electromagnetic radiation.
glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
>
>Let me quote from Panofsky and Phillips, _Classical Electricity and
>Magnetism_, 2nd ed., page 415-16.
>
>=========================================================================
> "In connection with this last point, there arises a question about
Be sure to note what that last point was.
>random scatterers. Consider a cubical volume whose edge is half a
>wavelength, and which contains many atoms or molecules. Even though the
>electrons may be distributed randomly within the cube, its radiation will
>be on average out of phase with that from a similar neighboring volume in
>a line of observation at right angles to the beam. The net observed
>scattering from many such volumes should then be zero from any angle,
>except, of course, the straightforward direction. Now these are the
>conditions on atmospheric scattering of visible light-- why, then, is
>Rayleigh's result in accord with observation?
> "The answer to this question has been given by Einstein and
>Smolukowski on the basis of a study of density fluctuations in the
>atmosphere. The number of scatterers per 'cell,' or pur cube of dimension
>[lambda]/2, is constant only on the average, and the fluctuations in this
>number are responsible for scattering.
Looks to me like you are confused by an unfortunate choice of words
at the end of that last sentence. They are responsible for the random
phases of the Rayleigh-scattered waves not resulting in total cancellation
of what you get for a single scatterer. The cause is, indeed, the
Rayleigh scattering by dipoles, which one recognizes by the polarization
of the scattered light.
>Therefore the total scattering is simply proportional to sum N_j=N, the
>total number of scattering centers, in agreement with Rayleigh's
>assumption.
That is, for the conditions of the atmosphere, the random fluctuations
in density mean that the scattered waves do not cancel and you can
sum them as Rayleigh assumed. They are not saying that refraction
by the density fluctuations is the cause of the scattering.
>The main point, I think, is that Rayleigh gets the correct result not
>because his assumptions were right, but because of an artifact of the
>statistics.
Note which assumptions were questioned: the additivity, not the
source and polarization of the scattered waves.
Paul Schlyter
In article <66ub41$6hh$1...@dismay.ucs.indiana.edu>,
Gregory Loren Hansen <glha...@copper.ucs.indiana.edu> wrote:
> In article <66tlpq$d...@electra.saaf.se>,
> Paul Schlyter <pau...@electra.saaf.se> wrote:
>>In article <66s87p$1v5$1...@dismay.ucs.indiana.edu>,
>>Gregory Loren Hansen <glha...@copper.ucs.indiana.edu> wrote:
>
>>>>> Why is our sky blue? I mean I know that the atmosphere scatters blue light.
>
>>> Light is scattered from density fluctuations in the air, which causes
>>> fluctuations in the dialectric constant. A single molecule in a vacuum
>>> visible light there's tens of thousands of molecules. It's those
>>> fluctuations in the dialectric constant that the light actually "sees"
>>> (i.e. changes in the index of refraction). The peak, for light from the
>>> wavelengths; the color we see isn't directly related to the energy in each
>>> wavelength that's hitting our eye.
>>
>>different mechanisms. Refraction is quite predictable, and follows
>>the laws of geometrical optics (this is why lenses and mirrors work),
>>while scattering is stocastic (you cannot predict in which direction
>>one single photon will scatter, only the statistical distribution
>>of the scattered photons).
>>
>>Refraction will yield its own set of phenomena, such as the apparent
>>flattening of the sun/moon when they are very near the horizon, the
>>apparent "raising" of the sun/moon at rise/set such that we still may
>>see them even though they "really" are below the horizon.
>
> You're thinking of refraction from a uniform medium, like from a prism.
index won't change in such a medium. There will be refraction only in a
non-uniform medium.
> I'm talking about scattering.
> And the only thing that scatters light is a change in the index of
> refraction; that's *the* optical property, the only thing that does
> anything at all with light.
at the molecular/atomic level.
> Except I'm saying there are too many molecules in a wavelength of air to
> treat it as scattering from individual molecules, it would transmit as if
> through glass. The scattering is from fluctuations in the air.
uniform air.
> I.e. put bubbles in your ice.
>
>>The blue color of the (clear) sky is due to scattering, not to
>>refraction.
>
> I said that. But you do you think the light scatters? You say molecules,
> I say density fluctuations. We get the same frequency dependence, anyway.
fluctuations.
>>This is why the blue color os so smoothly distributed,
>>as opposed to atmospheric-optical phenomena due to refraction and
>>reflection: raimbows, haloes, mirages, ....
>
> I thought rainbows were caused by scattering from water droplets, similar
> to the effect of a diffraction grating.
geometric optics can be applied.
If the water drops get smaller, diffraction enters the picture too: then
we will, immediately inside the primary rainbow, see a series of
supernumerary rainbows, alternating between bluish and reddish, several
times. At most, I've seen 4 such supernumerary rainbows inside the
primary rainbow.
If the water drops get even smaller, the diffraction effects start to
dominate: now the primary rainbow becomes invisible, and we only see
the supernumerary rainbows, wich colors so well mixed that this
rainbows appears colorless. This is the "white rainbow" -- it's a few
degrees smaller than the normal primary rainbow. I've seen it several
times, from airplanes, against the clouds below.
And if the water droplets get really small, the "white rainbow"
will shrink even more in size, and finally just appear as a "glory",
a few degrees wide, around the point directly opposite the Sun.
This is the glory, or anti-corona, which can be seen only if you
have clouds below you -- i.e. from a mountaintop, or an airplane.
The anti-corona is very often seen from airplanes.
Gregory Loren Hansen
In article <671l4h$k2k$1...@news.fsu.edu>,
Jim Carr <j...@ibms48.scri.fsu.edu> wrote:
>> "The answer to this question has been given by Einstein and
>>Smolukowski on the basis of a study of density fluctuations in the
>>atmosphere. The number of scatterers per 'cell,' or pur cube of dimension
>>[lambda]/2, is constant only on the average, and the fluctuations in this
>>number are responsible for scattering.
>
> Looks to me like you are confused by an unfortunate choice of words
> at the end of that last sentence. They are responsible for the random
> phases of the Rayleigh-scattered waves not resulting in total cancellation
> of what you get for a single scatterer. The cause is, indeed, the
> Rayleigh scattering by dipoles, which one recognizes by the polarization
> of the scattered light.
You're picking at words. OF COURSE it's ultimately due to dipole
radiation. Every optical effect is (neglecting those that aren't),
including diffraction, double-slit interference, refraction, and
reflection. But, at least on the elementary levels, you don't need to
analyze that phenomena on a dipole radiation level.
The point I was and am still trying to make is if there were no
fluctuations in the air, the sky wouldn't be blue. Dipole radiation is
ultimately responsible, but it is not a sufficient explanation. To
explain the color of the sky as Rayleigh scattering and then stop
there is misleading.
In quantum mechanics class, the professor used the color of the sky as an
example. He used fluctuations in the dialectric constant of the air and
Fermi's Golden Rule. He estimated the magnitude of the fluctuations
experimentally by considering a mirage on the road. We didn't need to go
to dipoles to get the right answer.
>>Therefore the total scattering is simply proportional to sum N_j=N, the
>>total number of scattering centers, in agreement with Rayleigh's
>>assumption.
>
> That is, for the conditions of the atmosphere, the random fluctuations
> in density mean that the scattered waves do not cancel and you can
> sum them as Rayleigh assumed. They are not saying that refraction
> by the density fluctuations is the cause of the scattering.
On a macroscopic level we usually talk about index of refraction, not
thirty thousand dipoles. It's the same thing, really. But as I said,
saying "The sky is blue because of Rayleigh scattering" is an incomplete
answer.
Gregory Loren Hansen
In article <671les$j...@electra.saaf.se>,
Paul Schlyter <pau...@electra.saaf.se> wrote:
>In article <66ub41$6hh$1...@dismay.ucs.indiana.edu>,
>Gregory Loren Hansen <glha...@copper.ucs.indiana.edu> wrote:
>>>Refraction will yield its own set of phenomena, such as the apparent
>>>flattening of the sun/moon when they are very near the horizon, the
>>>apparent "raising" of the sun/moon at rise/set such that we still may
>>>see them even though they "really" are below the horizon.
>>
>> You're thinking of refraction from a uniform medium, like from a prism.
>
>In any uniform medium there won't be any refraction, since the refraction
>index won't change in such a medium. There will be refraction only in a
>non-uniform medium.
Smooth, then. Or uniform on the order of a few wavelengths of light. At
that scale, the index of refraction of the air is a bit grainy.
>> I'm talking about scattering.
>
>Then you shouldn't be talking about density fluctuations....
In a macroscopic sense, that's what causes the scattering. If you want to
pick nits and look at the situation on a molecular level, the magnitude in
every direction but forward will average to zero because there are so darn
many randomly distributed dipole radiators. The reason enough of a
magnitude survives that we can see is because of the density fluctuations.
Those fluctuations are proportional to the number of molecules, which is
why Rayleigh got the right answer.
>> And the only thing that scatters light is a change in the index of
>> refraction; that's *the* optical property, the only thing that does
>> anything at all with light.
>
>The refractive index is a macroscopic property. Scattering happens
>at the molecular/atomic level.
There are tens of thousands of molecules in a sphere of the diameter of a
wavelength of visible light. Do the math. I've mentioned to Jim Carr
that the color of the sky can be analyzed using fluctuations in the
dialectric constant of the air and Fermi's Golden Rule, without going down
to the level of individual molecules.
>> Except I'm saying there are too many molecules in a wavelength of air to
>> treat it as scattering from individual molecules, it would transmit as if
>> through glass. The scattering is from fluctuations in the air.
>
>It's not -- there will be Rayleigh scattering even in pure and perfectly
>uniform air.
Yes. And the scattered wave in any direction will be the average of the
waves scattered by every individual molecule. And since the molecules are
randomly distributed and so close together, they'll average to just about
zero except in the forward direction.
Deep down at the heart of the physics, the sky is blue because of Rayleigh
scattering. But that is an incomplete answer.
>>>The blue color of the (clear) sky is due to scattering, not to
>>>refraction.
>>
>> I said that. But you do you think the light scatters? You say molecules,
>> I say density fluctuations. We get the same frequency dependence, anyway.
>
>I say you are wrong -- light scatters also in air with no density
>fluctuations.
Is the argument finished? Because if it's not, I'm going to ask you to
prove it.
Eric Flesch
On 13 Dec 1997 10:50:50 +0100, Paul Schlyter wrote:
>Novaya Zemlya resides above the arctic circle, which
>means it has midnight sun in the summer and "noon night" in the
>winter. The people there know quite well at what date in January or
>February the sun reappears above the horizon -- therefore a temporary
>premature "return of the sun" will be quite noticeable there.
Yes, although Novaya Zemlya never had any permanent population, even
before the Soviets nuked it to hell (they tested lots of H-bombs there
-- not underground either).
Not to take away from Paul's excellent dissertation on light
scattering. :-)
Eric Flesch
Phil Gibbs
In article <6728mb$eja$1...@dismay.ucs.indiana.edu>, Gregory Loren Hansen
<glha...@copper.ucs.indiana.edu> writes
>On a macroscopic level we usually talk about index of refraction, not
>thirty thousand dipoles. It's the same thing, really. But as I said,
>saying "The sky is blue because of Rayleigh scattering" is an incomplete
>answer.
Here is a quote from "Subtle is the Lord" Pais
"Smoluchowski was delighted. In a paper published in 1911, he spoke of
Einstein's contribution as 'a significant advance'. However, he had not
quite understood Einstein's argument. In an appendix to his 1911 paper
Smoluchowski mentioned that the blue of the sky is due to two factors:
scattering off molecules and scattering that results from density
fluctuations. Einstein objected by letter. There is one and only one
cause for scattering :'Reileigh [sic] treats a special case of our
problem, and the agreement between his final formula and my own is no
accident.' Shortly thereafter, Smolumchowski replied;'You are completely
right'"
crs
Gregory Loren Hansen wrote:
> The point I was and am still trying to make is if there were no
> fluctuations in the air, the sky wouldn't be blue. Dipole radiation is
> ultimately responsible, but it is not a sufficient explanation. To
> explain the color of the sky as Rayleigh scattering and then stop
> there is misleading.
>
> In quantum mechanics class, the professor used the color of the sky as an
> example. He used fluctuations in the dialectric constant of the air and
> Fermi's Golden Rule. He estimated the magnitude of the fluctuations
> experimentally by considering a mirage on the road. We didn't need to go
> to dipoles to get the right answer.
>
[snip]
> On a macroscopic level we usually talk about index of refraction, not
> thirty thousand dipoles. It's the same thing, really. But as I said,
> saying "The sky is blue because of Rayleigh scattering" is an incomplete
> answer.
I'm probably crazy to wade into a pissing contest like this one but
Gregory is probably right. For one thing, most of the Rayleigh
scattering arguments that suppose that light is scattering off
individual molecules assume that those molecules have a dipole moment.
Remember that N2 and O2 do not. Furthermore, their polarizabilities are
low - so induced dipoles from collisions or interactions with radiation
will be small. On the other hand, density fluctuations in the upper
atmosphere are just the size and magnetude to cause the phenomenon.
Here's a little experiment: the next time you're flying over land on a
clear day, look up and away from the sun - the sky will be bright blue.
You will be at about 35000 feet (11 km) so a good portion of the
atmosphere will be below you. Now look down and away from the sun.
Except for haze, the ground pretty much retains its color. If gas
molecules are doing the scattering, the dense part of the atmosphere
should appear more blue than the rarefied part. But that's not what
happens. On the other hand, (delta N/N), the magnatude of a density
fluctuation becomes more significant in a rarefied atmosphere than in a
dense atmosphere.
For a fine discussion which includes both the molecular theory as
originally presented by Rayleigh and the fluctuation theory, See
Jackson's "Calssical Electrodynamics," second edition, pp. 422-427. One
gets the Rayleigh (1/lambda^4) dependence in both cases so that as
Gregory pointed out, just mouthing the words "Rayleigh scattering" is an
incomplete answer.
Cheers,
Chuck Szmanda
chu...@ultranet.com
Jim Carr
>radiation. Every optical effect is (neglecting those that aren't),
>including diffraction, double-slit interference, refraction, and
>reflection. But, at least on the elementary levels, you don't need to
>analyze that phenomena on a dipole radiation level.
But I believe you do have to analyze this one that way, and the
textbook quotation (editted out of my reply here) says as much
in "The number of scatterers per 'cell,' or pur cube of dimension
[lambda]/2, is constant only on the average, and the fluctuations
in this number are responsible for scattering."
>The point I was and am still trying to make is if there were no
>fluctuations in the air, the sky wouldn't be blue.
I agree.
>Dipole radiation is
>ultimately responsible, but it is not a sufficient explanation.
But I think it is a necessary part of the explanation.
>In quantum mechanics class, the professor used the color of the sky as an
>example. He used fluctuations in the dialectric constant of the air and
>Fermi's Golden Rule. He estimated the magnitude of the fluctuations
>experimentally by considering a mirage on the road. We didn't need to go
>to dipoles to get the right answer.
What matrix element of what operator was used?
I don't see how you can get the polarization from just refraction.
Jim Carr
chu...@scientist.com writes:
>
>For one thing, most of the Rayleigh
>scattering arguments that suppose that light is scattering off
>individual molecules assume that those molecules have a dipole moment.
>Remember that N2 and O2 do not.
All that is required is that they have an electric dipole
polarizability, not a static dipole moment. You derive it
for a bound charge, and molecules have bound charges.
>Furthermore, their polarizabilities are
>low - so induced dipoles from collisions or interactions with radiation
>will be small.
It only has to be enough to re-radiate, producing scattering of
the incident radiation.
>Here's a little experiment: the next time you're flying over land on a
>clear day, look up and away from the sun - the sky will be bright blue.
It always looks very dark blue to me.
>If gas
>molecules are doing the scattering, the dense part of the atmosphere
>should appear more blue than the rarefied part.
Why? It should only be brighter. And it is.
Gregory Loren Hansen
In article <677050$oog$1...@news.fsu.edu>,
Jim Carr <j...@ibms48.scri.fsu.edu> wrote:
>glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
>>In quantum mechanics class, the professor used the color of the sky as an
>>example. He used fluctuations in the dialectric constant of the air and
>>Fermi's Golden Rule. He estimated the magnitude of the fluctuations
>>experimentally by considering a mirage on the road. We didn't need to go
>>to dipoles to get the right answer.
>
> What matrix element of what operator was used?
>
> I don't see how you can get the polarization from just refraction.
In quantum mechanics, the potential plays the same role that the index of
refraction does in optics. You get reflection of particles when there's
an abrupt change in the potential, you get reflection of light when
there's an abrupt change in the index of refraction, etc.
Let me look. He did this just before summer and we weren't to be tested
on it, so I didn't pay a lot of attention.
I guess it wasn't Fermi's Golden Rule, it was the Born approximation. He
started with
(del^2 + epsilon(r) omega^2/c^2) psi(r) = 0
_______
epsilon(r) = epsilon + delta epsilon(r)
_______
define k^2 = epsilon omega^2/c^2
(del^2 + k^2) psi(r) = -delat epsilon(r) omega^2/c^2
~ -k^2 delta epsilon(r)
_______
when epsilon is close to unity
Then do the Born approximation thing where the wave is equal to the
incident wave plus the scattered wave, I'll let you do the rest because
I'm sure you know your Born approximation and doing this kind of math in
ASCII sure is annoying. Except on the subject of delta epsilon,
"The fluctuations in epsilon are caused by random density fluctuations
of the air molecules on the scale of lambda, the wavelength of the light.
___
delta epsilon = (d epsilon/d rho) [rho(r') - rho(r')]
___
For V>>lambda cubed and q!=0 the mean density rho will not contribute.
N 3
rho(r') = [sum] delta (r' - r_j)
j=1
The scattering amplitude is
2 N
f (t) = (k /4pi)(d epsilon/d rho) [sum] exp(i q.r_j)
k j=1
So the angle comes from the dot product in the exponent. The magnitude of
the fluctuations was estimated experimentally by looking at mirages on the
road.
Jim Carr
Jim Carr <j...@ibms48.scri.fsu.edu> wrote:
|
| glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
| >In quantum mechanics class, the professor used the color of the sky as an
| >example. He used fluctuations in the dialectric constant of the air and
| >Fermi's Golden Rule. He estimated the magnitude of the fluctuations
| >experimentally by considering a mirage on the road. We didn't need to go
| >to dipoles to get the right answer.
|
| What matrix element of what operator was used?
|
| I don't see how you can get the polarization from just refraction.
glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
>
>In quantum mechanics, the potential plays the same role that the index of
>refraction does in optics. ... etc ...
>The scattering amplitude is
>
> 2 N
> f (t) = (k /4pi)(d epsilon/d rho) [sum] exp(i q.r_j)
> k j=1
>
>So the angle comes from the dot product in the exponent.
Which leads to refraction with no mention of a polarization effect.
Happy Holidays; I will wave when I pass by Bloomington. Feel free
to e-mail me a reply as well as post it, although it should last
until my return.
Gregory Loren Hansen
In article <67eac9$4nr$1...@news.fsu.edu>,
Jim Carr <j...@ibms48.scri.fsu.edu> wrote:
>glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
>>So the angle comes from the dot product in the exponent.
>
> Which leads to refraction with no mention of a polarization effect.
Yeah, I realized after I logged off that I didn't answer the question.
Classically speaking, you get partial polarization on reflection, maybe
you can work out the polarization of the sky from that. Quantum
mechanically, I'm not sure what to think of it. I'm sure that calculation
has a sum over polarization states hidden somewhere. You might have to go
all the way to dipoles to get that.
BUT... people usually ask "Why is the sky blue?", not "Why is the sky
polarized?"
> Happy Holidays; I will wave when I pass by Bloomington. Feel free
> to e-mail me a reply as well as post it, although it should last
> until my return.
I won't be here. I'm going home to Minnesota for the holidays. But I'll
be here in spirit to catch your wave.
Jim Carr
Jim Carr <j...@ibms48.scri.fsu.edu> wrote:
|
| glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
| >So the angle comes from the dot product in the exponent.
|
| Which leads to refraction with no mention of a polarization effect.
glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
>
>Classically speaking, you get partial polarization on reflection, maybe
>you can work out the polarization of the sky from that. Quantum
>mechanically, I'm not sure what to think of it. I'm sure that calculation
>has a sum over polarization states hidden somewhere. You might have to go
>all the way to dipoles to get that.
That is what I was thinking about in my comments: it is the sum on
polarization states that would cancel out along with the scattering
if not for the inhomogeneity.
>BUT... people usually ask "Why is the sky blue?", not "Why is the sky
>polarized?"
But the latter is the confirmation of Rayleigh scattering ... and
used by our Viking ancestors (guessing from name and Minnesota
relatives) to navigate.
crs
Jim Carr wrote:
>
> Jim Carr <j...@ibms48.scri.fsu.edu> wrote:
> |
> | glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
> | >So the angle comes from the dot product in the exponent.
> |
> | Which leads to refraction with no mention of a polarization effect.
>
> glha...@copper.ucs.indiana.edu (Gregory Loren Hansen) writes:
> >
> >Classically speaking, you get partial polarization on reflection, maybe
> >you can work out the polarization of the sky from that. Quantum
> >mechanically, I'm not sure what to think of it. I'm sure that calculation
> >has a sum over polarization states hidden somewhere. You might have to go
> >all the way to dipoles to get that.
>
> That is what I was thinking about in my comments: it is the sum on
> polarization states that would cancel out along with the scattering
> if not for the inhomogeneity.
>
> >BUT... people usually ask "Why is the sky blue?", not "Why is the sky
> >polarized?"
>
> But the latter is the confirmation of Rayleigh scattering ... and
> used by our Viking ancestors (guessing from name and Minnesota
> relatives) to navigate.
Just a question: Have you estimated the amount of polarization that
might obtain from randomly oriented scatterers much smaller than
lambda?
Actually, what has troubled me about the molecular scattering theories
is that the amount and brightness of blue in the sky can't be predicted
from the known polarizabilities of O2 and N2 (See Jackson). Additional
scattering from other sources must be invoked. After a little search, I
came across a review paper on this subject that may have at least a
partial answer to this question. Here's the reference:
Remarks on the theory of the blue of the sky
Richter, G., Ann. Phys. (Leipzig), 47(4). 325-39, (1990).
According to the abstract, scattering from molecules as well as
scattering from statistical fluctuations must be invoked if one is to
account fully for the observed color and intensity. I've ordered the
paper and, assuming I can still get through the German, I'll let you
know what I find.
Chuck Szmanda
chu...@ultranet.com