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robinpike
2007-Nov-12, 01:18 PM
On the recent post "How much do we know about mirrors, photons, and the mechanics of reflection", there seemed to be some contention as to whether we do, or do not, know how light is reflected from a surface, such as the partial reflection from a pane of glass. Ignoring the scattering of light by a surface, i.e. the random dispersion of light by surfaces, can someone, if we do know how light is reflected by the glass, explain how it works?

As mentioned in the other post, what seems to be a key part to the explanation, is how the light gets reflected (as opposed to scattered), so please include that bit in the explanation. Also I believe that the reflected light does not undergo any change in colour, so please can the explanation include why, for example, when light is reflected from silver, how the reflected light remains unchanged in colour?

And as a secondary (less important) set of questions, does the amount of light that is reflected (as opposed to dispersion) depend on the angle of incidence (and in what way)? And for a particular angle of incidence, is the amount of light reflected directly proportional to the atomic weights of the atoms in the substance (i.e. the heavier substances reflect more light), or what is the connection if any?

Sorry lots of questions, but at school (a long time ago) I never understood how all this works, and still don't! At school we were shown how to calculate what happens, rather than how it works.

Jim
2007-Nov-12, 03:19 PM
Moved to Q&A.

Jeff Root
2007-Nov-12, 05:19 PM
Hello, Robin!

We "know" how light interacts with matter to the extent that
the behavior of the light can be very accurately predicted by
theory in a wide range of circumstances. There is currently
no way to explain it simply.

The laws of optics describe some of the behaviors of light.
I expect that those laws are what you learned at school. They
include such things as the fact that the angle of reflection
equals the angle of incidence, and Snell's law, which relates
the angle of refraction to the angle of incidence at the
juncture of two different media.

Maxwell's equations, which define classical electrodynamics,
also describe some of the behaviors of light.

Quantum electrodynamics (or QED) is the modern theory which,
as far as I know, describes all behaviors of light in those
circumstances which can be adequately defined. I have not
studied QED, but I did read a nice little book about it by
physicist Richard Feynman titled "QED - The Strange Theory
of Light and Matter" (1985, Princeton University Press).
It is fairly easy to read (that was its reason for being)
and inexpensive in paperback. I strongly recommend that you
read this. It won't answer your questions, but it will come
as close to answering them as you are likely to want to get.

The material of the book was originally presented in four
lectures. Here is a nice paragraph from page 9:


What I am going to tell you about is what we teach our
physics students in the third or fourth year of graduate
school-- and you think I'm going to explain it to you so
you can understand it? No, you're not going to be able to
understand it. Why then, am I going to bother you with all
this? Why are you going to sit here all this time, when
you won't be able to understand what I am going to say?
It is my task to convince you not to turn away because you
don't understand it. You see, my physics students don't
understand it either. That is because I don't understand
it. Nobody does.
And a bit later on page 12:


I'm going to explain to you what the physisists are doing
when they are predicting how Nature will behave, but I'm
not going to teach you any tricks so you can do it
efficiently. You will discover that in order to make
any reasonable predictions with this new scheme of quantum
electrodynamics, you would have to make an awful lot of
little arrows on a piece of paper. It takes seven years--
four undergraduate and three graduate-- to train our physics
students to do that in a tricky, efficient way. That's where
we are going to skip seven years of education in physics:
By explaining quantum electrodynamics to you in terms of
what we are really doing, I hope you will be able to
understand it better than do some of the students!

I also have a copy of the October 1980 Scientific American
magazine. The cover story is 'The Causes of Color' by
Kurt Nassau. The brief abstract under the title says:
"They are diverse, but they all stem from the same root:
It is the electrons in matter, through their varied
responses to different wavelengths of light, that make
the world a many-colored place". A table in the article
lists various types of causes of color, with examples.
The caption of the table reads:


Causes of color are classified in 14 categories of five
broad types. All but one of the color-causing mechanisms
(vibrations of the atoms in molecules) can be traced to
changes in the state of the electrons in matter. Electronic
transitions are the most important causes of color because
the energy needed to excite an electron commonly falls in
the range that corresponds to visible wavelengths of light.
Get ahold of that issue if you can.

-- Jeff, in Minneapolis

robinpike
2007-Nov-12, 06:40 PM
Jeff, that's all good stuff - thanks, but am I right in thinking that quantum electrodynamics does not have in it, for example, the explanation as to how the angle of reflection is the same as the angle of incidence?

I guess what I am interested in is understanding how atoms (or at whatever lower level) seem to have two different types of processes going on when it comes to light meeting a surface of atoms: reflection and scattering.

For reflected light there seems to be two key questions: a) how the atoms cause the angle of reflection to be the same as the angle of incidence; b) how the colour of the light is not changed.

For scattered light it seems we do have an explanation, i.e. a) a single electron (or nucleus) 'absorbs' a single photon of light and at some point later that excited electron (or nucleus) emits the light back out in a random direction; b) when the light is absorbed and subsequently emitted, the excited electron (or nucleus) can emit the absorbed light in several stages, and thus emit the scattered light in (at least) two different colours of lower frequency to the colour of the original incident light.

For example, when 'white' light falls on polished gold, the scattered light is yellow and is scattered in all directions; whereas the reflected light remains 'white' and follows a reflected path.

So is it a case that reflected light and scattered light are caused by different mechanisms? Presumably qed shows us what is going on?

frankuitaalst
2007-Nov-12, 08:51 PM
There's an interesting link about reflection herunder :
http://www.spaceandmotion.com/Physics-Richard-Feynman-QED.htm
Quote :
For a mirror, the light waves are not really reflected. Instead, high energy / wave state electrons in the light source resonantly couple with the electrons in the mirror, causing these to move into a higher energy wave state. These electrons in the mirror then effectively become the light source, and resonantly couple with electrons in the detector, giving the appearance that light has gone from the source to the detector by 'reflecting' of the mirror.
End Quote
Reading this I think the resonanting source has the same frequency as the light which comes in .

G O R T
2007-Nov-13, 12:46 AM
I guess what I am interested in is understanding how atoms (or at whatever lower level) seem to have two different types of processes going on when it comes to light meeting a surface of atoms: reflection and scattering.

The terms are specular and diffuse reflection. Both are reflection but one keeps coherence. As light waves encounter a reflective surface they are reflected at angles relative to the plane formed by atoms in that area over a wavelength. If the surface is rough you get diffuse reflection. As the area is made smoother by polishing, the angles converge and specularity increases. When the surface is smooth enough that remaining divergent areas are well under a wavelength, you get a "mirror" finish. Your main question should be: why does light act only like a wave and reflect in accordance with alignment of wavelength sized areas instead of encountering individual atoms as a photon particle.



There's an interesting link about reflection herunder :
http://www.spaceandmotion.com/Physics-Richard-Feynman-QED.htm
Quote :
For a mirror, the light waves are not really reflected. Instead, high energy / wave state electrons in the light source resonantly couple with the electrons in the mirror, causing these to move into a higher energy wave state. These electrons in the mirror then effectively become the light source, and resonantly couple with electrons in the detector, giving the appearance that light has gone from the source to the detector by 'reflecting' of the mirror.
End Quote
Reading this I think the resonanting source has the same frequency as the light which comes in .

This is only a semantic change in explanation. We know photons are absorbed raising an atoms energy state, and then reemitted. "Resonantly couple" is an indication that photons have no travel time from their perpective, even though they do from ours. Wavelength only changes during propagation if the photon is sufficiently energetic for Compton scattering (normally X-rays and up) or absorbtion and reemittance as infrared.

robinpike
2007-Nov-13, 12:23 PM
The terms are specular and diffuse reflection. Both are reflection but one keeps coherence. As light waves encounter a reflective surface they are reflected at angles relative to the plane formed by atoms in that area over a wavelength. If the surface is rough you get diffuse reflection. As the area is made smoother by polishing, the angles converge and specularity increases. When the surface is smooth enough that remaining divergent areas are well under a wavelength, you get a "mirror" finish. Your main question should be: why does light act only like a wave and reflect in accordance with alignment of wavelength sized areas instead of encountering individual atoms as a photon particle.

Thanks G O R T. To summarise are you saying that all light falling on a surface which is then radiated out, is reflected light - which is either regarded as diffuse or specular depending on how smooth the surface is?

In that case, if specular and diffuse reflection are both caused by 'the same reflection process', then I don't understand why when, for example light is reflected from a flat smooth polished gold surface, the light that is reflected as in the 'mirror reflection' is unchanged in colour; and yet the other reflected light as in 'diffuse reflection' is changed in colour (to yellow, thus giving the gold its yellow colour)?

I don't understand where the mechanisms for these two different types of 'reflected light' are given / explained?

Jeff Root
2007-Nov-13, 12:27 PM
...am I right in thinking that quantum electrodynamics does not
have in it, for example, the explanation as to how the angle of
reflection is the same as the angle of incidence?
Answering that kind of question is exactly what QED does.

As I said:


Quantum electrodynamics (or QED) is the modern theory which,
as far as I know, describes all behaviors of light in those
circumstances which can be adequately defined.



I guess what I am interested in is understanding how atoms (or at
whatever lower level) seem to have two different types of processes
going on when it comes to light meeting a surface of atoms: reflection
and scattering.
If there is a college or university near you that has a physics degree
program, visit a bookstore on or near the campus which carries physics
textbooks. Ask for help in finding a textbook on QED. If the bookstore
clerk can't help you find what you want, call on a lecturer or professor
to ask for recommendations. You can phone the department office,
or just walk into the department office and ask the secretary to direct
you to a lecturer or physicist who can help you select a textbook on
quantum electrodynamics.



For example, when 'white' light falls on polished gold, the scattered light
is yellow and is scattered in all directions; whereas the reflected light
remains 'white' and follows a reflected path.

So is it a case that reflected light and scattered light are caused by
different mechanisms? Presumably qed shows us what is going on?
The table in the Scientific American article lists gold -- along with
silver, iron, copper, and brass as other examples -- as having color
caused by transitions in a material caused by energy bands, and in
these cases, in metallic conductors. The article has a simple,
generic graph illustrating this. The caption reads:


ELECTRONIC STRUCTURE OF A METAL is distinguished by an
essentially continuous band of allowed energy levels. The band is
filled from the ground state up to an energy level called the Fermi
level; all higher energy states are empty and can therefore accept
excited electrons. A consequence of this electronic configuration
is that all wavelengths of radiation can be absorbed, from the
infrared through the visible to the ultraviolet and beyond. A material
that absorbs light of all colors might be expected to appear black;
metals are not black because an excited electron can immediately
return to its original state by reemitting a quantum with the same
wavelength as the absorbed one. The metallic surface is therefore
highly reflective.
The article doesn't go into much detail but I think it is a good
overview of the range of known causes of color -- that is --
emission, absorption, transmission, and reflection of light.

-- Jeff, in Minneapolis

G O R T
2007-Nov-13, 01:21 PM
Gold is always gold though a strong reflection may seem to drown this out.


Wiki gives a simple explanation. (http://en.wikipedia.org/wiki/Plasmon)

robinpike
2007-Nov-13, 01:55 PM
I think I'm okay on the explanation / mechanism as to how substances have colour, it is the specular reflected light that I am having trouble with.


A material that absorbs light of all colors might be expected to appear black;
metals are not black because an excited electron can immediately
return to its original state by reemitting a quantum with the same
wavelength as the absorbed one. The metallic surface is therefore
highly reflective

On that basis, there are two things that I don't understand:

(1) Reflected light (as for a reflected image) appears to be able to have any colour reflected. When the electrons in say glass absorb photons from a source of all wavelengths, wouldn't many of the absorptions result in the electrons going to energy levels that are outside of their allowed quantum levels?

Or is it a case that if the light is absorbed / emitted quickly enough (i.e. re the uncertainty principle), it can do that forbidden energy jump?

(2) Even if the absorption / emission is allowed, I still don't see where the mechanism is for how that electron / photon coupling is able to emit the light in a direction that is at an angle of reflection equal to the angle of incidence of the whole array of atoms on the surface?

How is the mechanism of an individual photon interacting with an individual electron achieving that angle of reflection?

Spaceman Spiff
2007-Nov-13, 03:25 PM
There's an interesting link about reflection herunder :
http://www.spaceandmotion.com/Physics-Richard-Feynman-QED.htm
Quote :
For a mirror, the light waves are not really reflected. Instead, high energy / wave state electrons in the light source resonantly couple with the electrons in the mirror, causing these to move into a higher energy wave state. These electrons in the mirror then effectively become the light source, and resonantly couple with electrons in the detector, giving the appearance that light has gone from the source to the detector by 'reflecting' of the mirror.
End Quote
Reading this I think the resonanting source has the same frequency as the light which comes in .

While this site has some nice passages from Feynman's book on QED (search on Feynman, skip the rest) that are worth reading if you do not have the book, the model presented, "Spherical Standing Wave Structure of Matter", is not mainstream physics or maybe it is (?), sort of. Whoever this is seems to have dressed up this model to make all the correct predictions of QED (which it must, given the fantastic predictive power of QED). The major difference seems to be that the author of this model dismisses the photon concept, one that however appears to be a misconception in his/her mind (let's shoot the strawman) and which isn't considered in Quantum Field theory. Although I am not certain - I do not have the time to try to blow through the subterfuge and understand what the big deal is.

Feynman's book is a great buy and the guy is amazing in illuminating complicated concepts. It is worth your time. I suggest reading it instead.

robinpike
2007-Nov-13, 06:52 PM
So the theory of quantum electrodynamics gave the right answer - the middle of the mirror is the important part for reflection - but this correct result came out at the expense of believing that light reflects all over the mirror, and having to add a bunch of little arrows together whose sole purpose was to cancel out. All that might seem to you to be a waste of time - some silly game for mathematicians only. After all, it doesn't seem like real physics to have something there that only cancels out! (Richard Feynman, 1985).

Thanks Spaceman Spiff, that is a very good site, having the additional explanatory text works well alongside RF's easy way of explaining things.

RF goes on to explain how the hypothesis that the light is reflecting off the whole mirror is tested further:-


But let's suppose we carefully scrape the mirror away in those areas whose arrows have a bias in one direction. (Which is causing the waves to cancel out.) Isn't it wonderful, - you can take a piece of mirror where you didn't expect any reflection, scrape away part of it, and it reflects. (Richard Feynman, 1985)

Although the explanation for how the angle of reflection is the same as the angle of incidence seems to be asking the light to be doing something impossible, and yet further tests on that idea seem to confirm that the given explanation as being what happens.

So from that point of view, I am getting closer to understanding how QED explains reflection (as in a reflected image) - so many thanks for that.

So to completely tidy this one up, can someone just confirm what exactly the light is doing when it interacts with the whole mirror, presumably in the sense of how each photon interacts with the many many electrons simultaneously along the mirror's surface?

As that part is still unclear to me.

Ken G
2007-Nov-14, 03:41 PM
(1) Reflected light (as for a reflected image) appears to be able to have any colour reflected. When the electrons in say glass absorb photons from a source of all wavelengths, wouldn't many of the absorptions result in the electrons going to energy levels that are outside of their allowed quantum levels?Your question is a very deep one, and very insightful. Perhaps it would help to contrast these various things:
mirror
white piece of paper
colored piece of paper
gold bar.
The mirror is smooth on both the scale of the wavelength, so it specularly reflects, and on the scale of the resolution of your eye, so it can make images. The white piece of paper has the former property, but not the latter-- it diffusely reflects on the scale of your eye's resolution by specularly reflecting (and preserving color) on the scale of the wavelength. But a colored piece of paper changes white light into colored light, but that's by having a different reflectance at each color-- if you shine pure red light on a green piece of paper is it red or green? It's red, but just not nearly as bright as if you had shined green light. The gold bar is a combination of the mirror and the colored paper, the diffuse emission looks gold but only if you shine white light on it. That leads to your second question:

(2) Even if the absorption / emission is allowed, I still don't see where the mechanism is for how that electron / photon coupling is able to emit the light in a direction that is at an angle of reflection equal to the angle of incidence of the whole array of atoms on the surface?

How is the mechanism of an individual photon interacting with an individual electron achieving that angle of reflection?That is related to "interference" and the two-slit diffraction pattern. A photon does not do just one thing, even if it interacts with atoms-- it can do a lot of things, and what really happens requires adding up, and interfering, all those things. The place where you get constructive interference, not destructive, is any path that takes the least time between two points. The path of least time gives you Snell's law of refraction, and it also gives you angle of incidence equals angle of reflection. If you are in a race between two points, and you must touch a wall that is to one side before you can go to the destination, you run so as to have the angle of incidence equal angle of reflection-- to take least time.

Ken G
2007-Nov-15, 02:30 AM
And to add to that, the answer to why gold does not change the color when it specularly reflects, but does when it diffusely reflects, is that the surface of gold is very smooth. That means the only way to get diffuse reflection is to monkey with the phase of the light-- the phase of the waves that tell the photons where to go. If you don't change the phases, you get the destructive interference I was talking about, so angle of incidence equals angle of reflection. If you do monkey with the phase, then you are talking about processes that can also absorb the light-- so you get more diffuse scattering at some frequencies than at others, because you have different absorbing properties due to the structure of how the electrons are bound in the metal. Specular reflection won't do that because it doesn't alter the phases, so can't absorb light preferentially at some wavelengths. I don't know if one can view that as an uncertainty principle issue, perhaps one can-- it's true that the specular reflection is "immediate", whereas processes that can absorb some frequencies "take longer".

Spaceman Spiff
2007-Nov-15, 02:55 AM
The mirror is smooth on both the scale of the wavelength, so it specularly reflects, and on the scale of the resolution of your eye, so it can make images. The white piece of paper has the former property, but not the latter-- it diffusely reflects on the scale of your eye's resolution by specularly reflecting (and preserving color) on the scale of the wavelength.

What do you mean when you say above that piece of paper is smooth on the scale of the wavelength of visible light? Paper is a mat of macroscopic fibers, and light scatters amongst the fibers before making its way back out. I am apparently missing something subtle in what your trying to explain. And would you please define by what you mean by "your eye's resolution" (by that do you mean angular resolution?)
Thanks.


That is related to "interference" and the two-slit diffraction pattern. A photon does not do just one thing, even if it interacts with atoms-- it can do a lot of things, and what really happens requires adding up, and interfering, all those things. The place where you get constructive interference, not destructive, is any path that takes the least time between two points. The path of least time gives you Snell's law of refraction, and it also gives you angle of incidence equals angle of reflection. If you are in a race between two points, and you must touch a wall that is to one side before you can go to the destination, you run so as to have the angle of incidence equal angle of reflection-- to take least time.

This last bit is delightfully explained in Feynman's book, QED. Another reason for those interested to purchase it!

Jeff Root
2007-Nov-15, 09:23 AM
The mirror is smooth on both the scale of the wavelength, so it specularly
reflects, and on the scale of the resolution of your eye, so it can make
images. The white piece of paper has the former property, but not the
latter-- it diffusely reflects on the scale of your eye's resolution by
specularly reflecting (and preserving color) on the scale of the wavelength.

What do you mean when you say above that piece of paper is smooth on
the scale of the wavelength of visible light? Paper is a mat of macroscopic
fibers, and light scatters amongst the fibers before making its way back out.
I am apparently missing something subtle in what your trying to explain.
And would you please define by what you mean by "your eye's resolution"
(by that do you mean angular resolution?)
I believe I understand. Imagine that a sheet of white paper is made
of billions of very tiny, flat mirrors, with each mirror oriented randomly.
Each tiny mirror reflects specularly, but the mirrors are so tiny that
the human eye can't resolve them -- it only sees a hodgepodge of
reflections coming about equally from all parts of the sheet. In the
case of white paper, the "mirrors" are microscopic, but still larger than
the wavelengths of visible light. If the "mirrors" are about the size
of wavelengths of green light, you get a blue-tinted reflection. This
is, I think, the mechanism causing the blue color of the sky, the red
of sunsets, and the colors of moonstones and star sapphires. (How
astronomical!)

-- Jeff, in Minneapolis

robinpike
2007-Nov-15, 12:33 PM
Okay I'm beginning to understand all this a bit better now, so many thanks for all the posts.

Just to finish off, can anyone give details on whether reflected light (whether diffuse or specular) gets its frequency changed with regards to Doppler shift effects?

I'm thinking if the reflecting surface is moving, say away from the original light source, does this have any effect on the frequency of the reflected light?

My initial thoughts are that it shouldn't make any difference to the frequency of the reflected light. But then, how do those radar speed guns work?

Ken G
2007-Nov-15, 04:26 PM
What do you mean when you say above that piece of paper is smooth on the scale of the wavelength of visible light? Paper is a mat of macroscopic fibers, and light scatters amongst the fibers before making its way back out. I'm assuming the fibers are large compared to the wavelength, and reflect specularly, like a bunch of fiber-shaped mirrors. On larger scales, that doesn't look like a mirror, because your eyes can't resolve the fibers. Fibers might be doing something different, for all I know (multiple reflections between them would further complicate the picture, but it would be the same idea), I'm just drawing the distinction between the apparent whiteness of rough mirrors whose scale you cannot resolve (angular resolution, yes).

Spaceman Spiff
2007-Nov-15, 06:17 PM
I believe I understand. Imagine that a sheet of white paper is made
of billions of very tiny, flat mirrors, with each mirror oriented randomly.
Each tiny mirror reflects specularly, but the mirrors are so tiny that
the human eye can't resolve them -- it only sees a hodgepodge of
reflections coming about equally from all parts of the sheet. In the
case of white paper, the "mirrors" are microscopic, but still larger than
the wavelengths of visible light.

That much seems to jive with what Ken G is saying above but...


If the "mirrors" are about the size
of wavelengths of green light, you get a blue-tinted reflection. This
is, I think, the mechanism causing the blue color of the sky, the red
of sunsets, and the colors of moonstones and star sapphires. (How
astronomical!)

....cannot be quite right. Molecular scattering and molecular absorption (or for that matter the absorption and scattering of light within solids) are definitely not the same thing. The N2 and O2 molecules in our atmosphere are most certainly not absorbing visible light, selectively or otherwise, whereas the colors you see in gemstones are due to some combination of selective scattering and absorption.

I am looking at a reference that shows the absorption spectrum of an emerald. There are two big fat bands centered at 400 and 600 nm (due to the presence of chromimum mixed with the beryl matrix), with a minimum in absorption between them at 500 nm (our eyes call that "green" light).

Jeff Root
2007-Nov-16, 09:15 AM
If the "mirrors" are about the size of wavelengths of green light, you
get a blue-tinted reflection. This is, I think, the mechanism causing
the blue color of the sky, the red of sunsets, and the colors of
moonstones and star sapphires. (How astronomical!)
....cannot be quite right. Molecular scattering and molecular absorption
(or for that matter the absorption and scattering of light within solids)
are definitely not the same thing. The N2 and O2 molecules in our
atmosphere are most certainly not absorbing visible light, selectively
or otherwise...
I don't know, and perhaps I haven't thought it through adequately,
but what I said still seems okay to me. For the case of sunlight
passing through air, the molecules act like very tiny mirrors about
the size of the wavelength of green light, so that red and yellow go
right past, while blue is reflected, scattering blue in all directions.
When the Sun is low in the sky, the long column of air also reflects
enough of the green and yellow to allow only red to get through.

The mechanism producing the color of the blue sky and of the red
sun at sunset is scattering of shorter wavelengths. The Scientific
American article says that the bluish tint of a glass of diluted,
un-homogenized milk, and the reddish tint of a light beam passing
through the milk are caused by scattering, the bluish color of
moonstone is almost certainly is caused by scattering, and the
pattern in star sapphire is caused by scattering.

However, the article also says that the scattering is caused by
diffraction, not reflection, so you are right, too. No tiny mirrors.



whereas the colors you see in gemstones are due to some combination
of selective scattering and absorption.

I am looking at a reference that shows the absorption spectrum of an
emerald. There are two big fat bands centered at 400 and 600 nm
(due to the presence of chromimum mixed with the beryl matrix), with
a minimum in absorption between them at 500 nm (our eyes call that
"green" light).
Here's the table in the Sci Am article:


Electronic transitions in free atoms and ions;
Vibrational transitions in molecules:
Electronic excitations: - Incandescence, flames, arcs, sparks,
lightning, gas discharges, some lasers.
Vibrations - Blue-green tin of pure water and ice.
Crystal-field colors:
Transition-metal compounds - Turquoise, most pigments, some lasers,
some phosphors, some fluorescence.
Transition-metal impurities - Ruby, emerald, red sandstone, some
lasers, some fluorescence.
Color centers - Amethyst, desert amethyst glass,
smoky quartz, some fluorescence.
Transitions between molecular orbitals:
Charge transfer - Blue sapphire, magnetite.
Conjugated bonds - Organic dyes, most plant and animal
colors, fireflies, lapis lazuli,
dye lasers, some florescence.
Transitions in materials having energy bands:
Metallic conductors - Copper, silver, gold, iron, brass.
Pure semiconductors - Silicon, galena, cinnebar, diamond.
Doped semiconductors - Blue diamond, yellow diamond,
light-emitting diodes, semiconductor
lasers, some phosphors.
Geometrical and physical optics:
Dispersive refraction - The rainbow, "fire" in gemstones,
chromatic aberration.
Scattering - Blue of the sky, red of sunsets,
moonstone, star sapphire.
Interference - Oil film on water, lens coatings,
some insect colors.
Diffraction grating - Opal, liquid crystals,
some insect colors.
My hand hurts from holding the magazine up. :)

-- Jeff, in Minneapolis

robinpike
2007-Nov-16, 12:48 PM
If I think of this problem in terms of light reflecting from a moving mirror and the light as wave peaks, then the Doppler Shift that occurs to the reflected light could be thought of as follows:

As a wave peak of the light hits the mirror and reflects back, by the time the next wave peak hits the mirror, the mirror has moved away a bit, so not only does it reach the mirror a bit later, but it has slightly further to go on the return journey, so the peaks are further apart. Since light always travels with the same speed, peaks further apart means lower frequency.

I can understand that explanation until I add another, second mirror moving with the first mirror so as to bounce the reflected red-shifted light between the two.

So for example, a mirror is moving away from a source of ‘green’ light. As the ‘green’ light moves pass the mirror, some of it hits the mirror and is reflected back red-shifted, and therefore that reflected light becomes ‘red’ light. This ‘red’ light meets the second mirror, which is moving with the first mirror, but since this mirror meets the light head-on, this mirror blue-shifts the ‘red’ light back to ‘green’ light.

So far so good.

The problem that I’m having here is that both mirrors (and any light detectors placed at each mirror) should see the light bouncing between the two of them as the same colour, which presumably in this case should be ‘red’ light. But I keep ending up with the result that if the first mirror sees the ‘green’ light as ‘red’ light, then following the same process, the second mirror will see the ‘red’ light as ‘green’ light?

Where am I going wrong here?

robinpike
2007-Nov-16, 02:41 PM
As I'm not sure what I'm getting wrong here, perhaps if we compare a similar set up with sound waves in air, it might help, even though they are not the same.

Suppose a person is on an open railway cart moving away from a stationary source of sound, say a whistle. They will hear the note of the whistle at a lower frequency than someone not moving, ie they hear the note of the sound Doppler red-shifted.

If the person on the open railway cart holds a convex metal dish to reflect the sound back to a person sitting at the back of the cart, what frequency will that other person hear the sound at? Presumably the frequency of the note would be Doppler shifted back up again?

Now, in that example, the wave, ie the sound, is traveling in air and the people on the railway cart are moving with respect to the air, so it is not an equivalent comparison to what light is, or how light behaves.

What I don't seem to be able to get to grips with is a consistent understanding of what the light is doing / what is happening to the light as it is first red-shifted, and then bounced between the two mirrors?

I know that what we see is the initial light being red-shifted. And I also know that what we see as the light bounces between the two mirrors, is that the light does not change frequency. What I cannot seem to get a grasp of is the explanation that explains the Doppler shift and keeps the light at one frequency as it bounces between the mirrors?

Jeff Root
2007-Nov-16, 02:51 PM
Robin,

Light detectors attached to both mirrors will see the same amount
of redshift. Instead of thinking of the mirrors as moving away from
the lightsource, imagine the lightsource moving away from the two
mirrors. The situations are indistinguishable.

If you want to consider the mirrors as moving, then consider that
the frequency of peaks hitting the two mirrors has to be the same.
If twelve peaks hit the second mirror for every ten hitting the first,
there would be two peaks materializing from nowhere for every ten
that hit the first mirror.

-- Jeff, in Minneapolis

Jeff Root
2007-Nov-16, 02:56 PM
My previous post was written before your most recent post.

What we are talking about now is special relativity, and the
answers to your questions are provided by special relativity.
There are probably 50 to 100 threads on BAUT covering this
topic and the specific question you are asking. I'm not
particularly good at finding them, though...

-- Jeff, in Minneapolis

Jeff Root
2007-Nov-16, 03:04 PM
In the relativity way of looking at these situations, the light is not
affected by the motion of the source, the mirrors, or the observers.
Instead, it is the relationship between the light and the mirrors and
observers which is affected. An observer moving away from a light
source sees the light redshifted. An observer sees the light from a
source which is moving away from him redshifted. In both cases,
the light from the source is unchanged; it is only the relationship
between the observer and the wavecrests which changes.

-- Jeff, in Minneapolis

robinpike
2007-Nov-16, 06:17 PM
Jeff, you're right, this is really just a question in relativity, which will have been covered many times before. So many thanks to everyone for your help on this, I'll now go and brush up on my relativity.

Ken G
2007-Nov-19, 10:56 PM
The Scientific
American article says that the bluish tint of a glass of diluted,
un-homogenized milk, and the reddish tint of a light beam passing
through the milk are caused by scattering, the bluish color of
moonstone is almost certainly is caused by scattering, and the
pattern in star sapphire is caused by scattering.

However, the article also says that the scattering is caused by
diffraction, not reflection, so you are right, too.

That does not sound like the standard explanation, maybe the Scientific American article is saying the standard explanation is wrong? That would surprise me, but be very interesting. The standard explanation is not diffraction, it is "Rayleigh scattering", which says that when the energy of the photons is way below anything that can resonate with nitrogen molecules, they have to content themselves with "tickling" the bound electron at a frequency that is so slow compared to the restoring force that the electron is a kind of constant state of force balance. What that means is that the amplitude of the oscillation depends only on the electric field, and not on the frequency at all. What that means is that if red light and blue light have approximately the same strength fields (so same energy flux too), the electrons (classically) oscillate with the same amplitude. But the blue frequencies are faster, so the oscillation is faster, and this by itself makes the energy radiated scale like frequency to the 4th power. Note that this holds at all frequencies much smaller than resonant ones-- there is no need for there to be any break between green and red colors. Indeed, the normal explanation includes no such break, the frequencies are way below that.