How Does Light Behave When It Passes Through Different Mediums?

[whydoyouwanttoknow]

Ok, I don't want no equations and stuff please.

When light travels through a medium the photons are absorbed by atoms in the medium and then a photon is given off by the atom of similar energy, etc, this is correct?

If this is so does that mean when I look out the window that I'm not actually seeing the light given off by the sun but light from the window, it's new light?

 

[HallsofIvy]

Define "new light"! Light is light.

 

[whydoyouwanttoknow]

Define "new light"! Light is light.

Well is it counted as the light that came from the other side of the window or is it counted as light from the atoms in the glass?

 

[zoobyshoe]

When light is reflected, it is absorbed and re-emitted. Ivan Seeking told me last year that no one can say if the reflected light is exactly the same energy that was absorbed. Reflected light would be from "the atoms in the glass" as you put it.

I believe the same can be said of transmitted light. This is why light seems to be going less than the speed of light through glass, and other transparent stuff. It has to take a spin around various atoms that happen to absorb it. It is actually still doing that at c, the speed of light. It takes longer to get from point A to point B because it has to do a lot of going around in circles along the way.

 

[Claude Bile]

I believe the same can be said of transmitted light. This is why light seems to be going less than the speed of light through glass, and other transparent stuff. It has to take a spin around various atoms that happen to absorb it. It is actually still doing that at c, the speed of light. It takes longer to get from point A to point B because it has to do a lot of going around in circles along the way.

I cannot overstate how wrong this is. Zoobyshoe, this is not the place for speculation.

I will stress that light is not absorbed when traveling through a transparent medium, there are no electrons being promoted into higher atomic orbitals. If a beam of light were absorbed in a medium, it would not be transmitted through that medium.

To understand what is happening when light passes through a medium, you need to understand what polarisation is. When one applies an electric field to an atom, the electric field causes the electron(s) to be slightly displaced from equilibrium, which forms an electric dipole. This pertubation can propagate through a medium, much the same way as if you jiggle a length of rope at one end, the pertubation (the jiggle), propagates down the length of the string. The energy and momentum of the incident light is carried in this wave, so essentially the light is, for all intents and purposes, traveling through the medium.

Why does light propagate slower in a medium than in a vacuum? As a perturbed atom passes the pertubation on to the nest atom, there is a slight shift in the phase of the oscillation of the two atoms. The culmination of these phase shifts manifests itself as a reduced velocity.

In short, the energy and momentum flung out by the sun in the form of visible radiation passes through the window. Although the window (and the atmosphere, for that matter) may nudge and tickle it a little bit, the packets of energy and momentum that we call photons that hit your eyes, for all intents and purposes originated from the sun.

Claude.

 

[zoobyshoe]

I cannot overstate how wrong this is. Zoobyshoe, this is not the place for speculation.

I can accept that this is wrong. However, it isn't speculation on my part. It's how I read a mentor explain it in a thread here last year.

 

[GENIERE]

As I understand the process, it IS one of absorbtion and re-emission. I'm not sure what Claude means by transparency, since that would be a function of the energy of the incident photon and the composition of the medium.

"Why does light propagate slower in a medium than in a vacuum?

Light can only propagate in a vacuum, else it is absorbed or undergoes some type of scattering event

 

[marlon]

Light can only propagate in a vacuum, else it is absorbed or undergoes some type of scattering event

or absorption event...you are correct.

Besides, if anyone ever wondered why leaves are green or why the sky is blue, check out my journal (the first page you see, "on the physics of colours").
Besides, don't forget that photons have alwys the same speed but an EM-wave has variable speed when passing through a medium : the wave can be slowed down for example. The clue is to look at this traffic analogy : suppose you have a car that has constant speed and travels a 100m road. There are three stopping lights on that road so the car will stop three times. The netto-effect will be that the car will take more time to finish the 100m and in this way : the total velocity is lowered, you see ? The stopping event can be seen as the absortion of a photon by an atom of the medium. Ofcourse, the car is the photon

regards

marlon

 

[DaveC426913]

Ok, I don't want no equations and stuff please.

When light travels through a medium the photons are absorbed by atoms in the medium and then a photon is given off by the atom of similar energy, etc, this is correct?

If this is so does that mean when I look out the window that I'm not actually seeing the light given off by the sun but light from the window, it's new light?

I'm going to go out on a limb and restate the question.

Of the light that reaches me after passing through a window, what percent of the photons have passed directly through the glass without being absorbed and re-emitted by at least one atom, and what percent of photons have interacted? (A ballpark number is sufficient.)

(As I ask this question, I realize that the photons have to make their way through a hundred miles of air too. The number of photons reaching me that actually left the Sun, is a pretty big zero.)

 

[whydoyouwanttoknow]

I'm going to go out on a limb and restate the question.

Of the light that reaches me after passing through a window, what percent of the photons have passed directly through the glass without being absorbed and re-emitted by at least one atom, and what percent of photons have interacted? (A ballpark number is sufficient.)

(As I ask this question, I realize that the photons have to make their way through a hundred miles of air too. The number of photons reaching me that actually left the Sun, is a pretty big zero.)

Yeah. So are they though counted as new photons or does physics count them as the photons that came from the sun?

 

[zoobyshoe]

Let's look more closely at Claude-Bile's explanation. I don't get the impression from peoples subsequent posts it's been generally understood:

I will stress that light is not absorbed when traveling through a transparent medium, there are no electrons being promoted into higher atomic orbitals.

Claude is stressing that the photon is not transduced into a more energetic electron in the case of transmission.

If a beam of light were absorbed in a medium, it would not be transmitted through that medium.

Meaning, I think, you could have many things as a result of absorption; (reflection ard the photoelectric effect are two I can think of) but you wouldn't get transmission.

To understand what is happening when light passes through a medium, you need to understand what polarisation is. When one applies an electric field to an atom, the electric field causes the electron(s) to be slightly displaced from equilibrium, which forms an electric dipole.

This is where Claude begins to explain a very different mechanism whereby the energy from the incoming photon is transmitted through the medium, a mechanism that isn't the same as absorbtion and re-emission at all. I hope everyone understands the imbalance of charge that he has described: when an atom is subject to an electric field the electrons in that atom are affected by the field. Specifically, the electrons are prevented from maintaining equilibrium (equilibrium with regard to the other electrons and nuclei of the atoms they're a part of, I assume). With its electrons shifted out of place, the atom becomes more positive or negative on one side or the other. It has become polarized, that is, electrically positive and negative poles have been formed in the atom. Whereas before it was electrically homogenous all around, now it is a dipole: positive on one side, negative on the other. (Claude may want to confirm or refute my paraphrase. I hope I understood it accurately.)

This pertubation can propagate through a medium, much the same way as if you jiggle a length of rope at one end, the pertubation (the jiggle), propagates down the length of the string.

If I understand what Claude has said so far, let me suggest that the rope analogy could be replaced by the better one of billiard balls lined up in a row, all touching each other. Strike one at the end with a force whose vector is pretty much in the same direction as the line of balls, and the energy of the blow will propagate down the line of balls as each strikes (perturbes) the next in line. (Subject to Claude's approval.)

The explanation Claude is giving is that when the first atom becomes polarized (by the incident photon, presumably) the electrical imbalance has a direct and imediate affect on the next atom in line: it finds itself subjected to an electric field that is more positive or negative than equilibrium, and it, too, forms a dipole in reaction, which only serves to present an electrically unbalanced field to the next atom, and so on.

The energy and momentum of the incident light is carried in this wave, so essentially the light is, for all intents and purposes, traveling through the medium.

Here, I want to interpolate a reaction I had when reading this: based on your description it doesn't seem accurate to call what is passing through the medium a "photon". To me the word designates a specific form of energy and momentum, in order, among other things, to distinguish it from other forms. I think it would be more accurate to say the photon has been transduced to a different form of energy for the duration of travel through the medium.

Why does light propagate slower in a medium than in a vacuum? As a perturbed atom passes the pertubation on to the nest atom, there is a slight shift in the phase of the oscillation of the two atoms. The culmination of these phase shifts manifests itself as a reduced velocity.

I would be interested in reading an expansion of this explanation. "Phase of oscillation" and what would constitute a shift of that phase, aren't clear enough to me that I can understand how a reduction in velocity would result. (I never bothered to dig deeper into the explanation that was given last year - your sounds much more promising.)

In short, the energy and momentum flung out by the sun in the form of visible radiation passes through the window. Although the window (and the atmosphere, for that matter) may nudge and tickle it a little bit, the packets of energy and momentum that we call photons that hit your eyes, for all intents and purposes originated from the sun.

Again, while the quantity of energy and momentum would be conserved, I think that calling it a photon while it is propagating through the medium doesn't seem accurate. Passing through the atmosphere or window, seems, from your description, to be doing more than nudging and tickling it a bit. The latter sounds too much like a pin ball making it intact through a field of pins sticking up that divert it back and forth a bit. The traveling dipole, or phase shift, seems different enough to me from a photon to merit a different name, although I don't know what would be most appropriate or accurate.

Anyway, I thought your careful post deserved more attention. The mechanism you described is clearly different from the common absorbsion and re-emission of a photon by an electron that I had stuck in my head.

 

[Claude Bile]

I can accept that this is wrong. However, it isn't speculation on my part. It's how I read a mentor explain it in a thread here last year.

Okay, fair enough, but..

I believe the same can be said of transmitted light.

The whole 'I believe' part was the phrase that I didn't like. Anyway, with regard to your next post..

I think you understood my earlier post quite well. In terms of energy propagation, the line of billiard balls is ok, though mathematically, one describes light in a medium using the (classical) wave-equation, which is why I prefer the string.

The physics of this is encapsulated by the Kramers-Kronig relations, which essentially treat the electron/atom system as a driven oscillator. If you are familiar with the physics of driven oscillators, the form of the equations will be very familiar and easy to understand and interpret.

In a driven oscillator, the motion of the body undergoing oscillation is out of phase with the driving force when one is not on resonance. In the case of transmission through a medium, it is a necessary condition that the driving force not be on a resonance, otherwise absorption will result. Now consider the case where the motion of one electron (A) is driving the motion of the next (B). Electron B is out of phase with electron A, i.e. there is a net phase shift between A and B. Now consider a medium as a long path of electrons, the accumulated phase shift results in a slowed velocity.

The electric permittivity of the medium affects the magnitude of the phase shift from one electron to another. Hence the relationship between refractive index and the velocity of light through a medium.

It is clear that you understood what I was saying, because you are asking the right questions, so to speak. Is a photon in a medium still a photon? Ask yourself, what is a photon? It is an oscillation in an electromagnetic field. What is this pertubation wave? It too, is an oscillation in an electromagnetic field. Essentially, the photon is still considered a photon inside a medium because all its properties are preserved (I won't get started on non-linear media).

You obviously have a grasp of this level of physics, I encourage you to read a little literature on the topic to gain a deeper understanding. Most graduate level books on optics or optoelectronics should mention the Kramers-Kronig relations. Also, most books on non-linear optics should deal with this topic thoroughly.

Regards,
Claude.

 

[Claude Bile]

Light can only propagate in a vacuum, else it is absorbed or undergoes some type of scattering event

Are you saying that photons can't be transmitted through any media? I think there is a misunderstanding of the processes of scattering and absorption here. While there will be some absorption and some scattering, the vast majority of photons through, say a 1m length of fibre will not undergo either of these processes.

(The probability of light being absorbed and reemitted or scattered in such a way that the photon is guided by the optic fibre again is very small).

or absorption event...you are correct.

Besides, if anyone ever wondered why leaves are green or why the sky is blue, check out my journal (the first page you see, "on the physics of colours").
Besides, don't forget that photons have alwys the same speed but an EM-wave has variable speed when passing through a medium : the wave can be slowed down for example. The clue is to look at this traffic analogy : suppose you have a car that has constant speed and travels a 100m road. There are three stopping lights on that road so the car will stop three times. The netto-effect will be that the car will take more time to finish the 100m and in this way : the total velocity is lowered, you see ? The stopping event can be seen as the absortion of a photon by an atom of the medium. Ofcourse, the car is the photon

Marlon, your explanation ignores one very important property of abosrption and spontaneous emission; The radiation must be isotropic in that the emitted photon has an equal probability of traveling in all directions. The actualy probability of an spontaneously emitted photon traveling in the same direction as the incident (absorbed) photon is infinitesimally small.

Yeah. So are they though counted as new photons or does physics count them as the photons that came from the sun?

We can infer properties of the sun by looking at its photons (spectrum), since the properties of the majority of photons coming directly from the sun have not changed. Helium was first discovered on the sun in this way by looking at its absorption spectrum (hence the name Helium, after Helios the greek sun god).

To answer your question, the photons, for all intents and purposes came from the sun.

Claude.

 

[T@P]

maybe this question can help clarify the discussion: how does a light filter work? (say a green piece of plastic, essentially a green filter)

does the green filter 'block' all light that isn't in the same wavelength/frequency as green light or is it more molecular, i.e. do the atoms in the filter release green light when light is shined on them? (the second idea sounds more like something flourescent)
anyway if you answer the question about the filter then consider a regular glass or plastic as a filter that filters nothing.

see no messy arguments this way :)

 

[GENIERE]

…In a driven oscillator, the motion of the body undergoing oscillation is out of phase with the driving force when one is not on resonance. In the case of transmission through a medium, it is a necessary condition that the driving force not be on a resonance, otherwise absorption will result. Now consider the case where the motion of one electron (A) is driving the motion of the next (B). Electron B is out of phase with electron A, i.e. there is a net phase shift between A and B. Now consider a medium as a long path of electrons, the accumulated phase shift results in a slowed velocity.

OK Claude I think I follow your explanation re: “driven oscillator”, but I would consider it to be an absorption event wherein there must be a communication of energy states between the atom and photon. This communication would be via virtual photons and is not instantaneous (virtual photons travel at C). Thus the atom/photon are briefly coupled delaying the photons propagation but, since quantum states are not altered, no direction change would take place. If an orbital electron absorbed the photon, subsequent re-emission would be directionally random (blue sky phenomena).

Re: electrons (A,B) above. Are these free electrons?

 

[Claude Bile]

The explanation I put forward is from a classical standpoint, to be honest, I do not have enough experience with QM to put forward a strong explanation from that perspective.

The electrons are bound, they are subject to a restoring force.

Claude.

 

[McQueen]

I was directed to this thread by Claude Bile and after reading through the posts I can only Conclude that while Claude accurately states the classical view of light transmission through glass , today this view is outdated . Also that Zoobyshoe should not have been so hasty in abandoning his mentor’s earlier statements. The following is an account of the modern view of light:
“When light enters a transparent medium, the ultraviolet photons and infrared photons are absorbed while the frequencies of visible light are transmitted. In the case of UV light, the electrons in the medium to begin to resonate, or vibrate with the influx of energy. Those vibrating electrons that strike neighboring atoms, release or transfer much of their vibrational energy in the form of heat. With IR, entire atoms (not just electrons) begin to vibrate, subsequently generating heat.
Those electrons which are free to vibrate without striking neighboring atoms complete two quantum mechanical processes: excitation and de-excitation. Excitation occurs when a ground state electron absorbs a photon and jumps up to a higher, unstable energy level. Photons are bundles of radiant energy that represent the particle nature of light. The amount of energy present in a photon is calculated with the equation E = hf where f is the frequency of the light wave and h is Planck's constant, 6.64 x 10^-34 J sec. When the electron falls back to its ground state it releases a photon. The energy of the released photon exactly match the difference in the electron energy states and the energy of the initially absorbed electron. This process is called de-excitation. The photon emitted is then free to travel at 3 x 10^8 m/sec until it is again absorbed by another electron. There is no energy lost in the process.
The closer the energy of the photon is to a difference in the fundamental energy states of the atom, the more interaction takes place. Since the energy of the emitted photon exactly equals the energy of the absorbed photon, the frequency of the photons/light does not change. However, the time delay caused by this absorption/readmission process increases the time required for a photon to travel through the medium and therefore results in a slower average speed of light in that particular medium. The amount of time delay is evidenced by the optically dense medium's index of refraction; the greater the value of n, the more interaction and the greater the amount of refraction. “

Finally in conclusion I would have to state that whatdoyouwanttoknow was correct in his conclusions that none of the original photons emitted by the sun reach the observer after passing through the glass. At the same time it should be remembered that photons possessing identical frequencies are identical and thus the photons that eventually arrive are indistinguishable from those that set out.

 

[zoobyshoe]

McQueen,
You are explaining why UV light is not transmitted through glass, and why IR light heats glass up.

This does nothing to invalidate Claude's explanation of the transmission of all the light that goes "directly" through.
I don't believe what he said is outdated at all. He explained it on the quantum level quite nicely. He didn't explain it with regard to QED because QED doesn't explain it at all, just calculates probabilities.

 

[timeformation]

just wanted to add, think of a twodimensionalperson traveling ina three dimensionalworld. True, he still has all that he needs, in the sense ofa two dimensional world to travel in, butnow he has a third vector affecting him. Light is like a screw. but this screw has no threads along it. Now imagine the light is the person walking in his twodimensional world. As he travels and ius affected, then he gets wounded. His battle scars are the threads. Now, to a man observing the two dimensional person, he can't see the shaft which is the real portion of the screw, he merely detects the threads. Remember a three dimensional person sees only three dimensional things. So the threads, or wounds of the screw, are detected as wavelengths orbitting around a core ( shaft) the wavelength detected is merely the threads being carved by certain materials ( ie atoms) hence why hydrogen gives off a certain color, and magnesium another...

keep inmind, the same column in periodic table, has similar light emission patterns, but has added protions to it..

better andsiumple explanation once you get your head around it

adam

 

[zoobyshoe]

I think you understood my earlier post quite well. In terms of energy propagation, the line of billiard balls is ok, though mathematically, one describes light in a medium using the (classical) wave-equation, which is why I prefer the string.

OK. I understand your preference for the string.

The physics of this is encapsulated by the Kramers-Kronig relations,

I googled the KK relations and found they are in the area of calculus. Being a math basket case, I cannot penetrate into that land.

which essentially treat the electron/atom system as a driven oscillator. If you are familiar with the physics of driven oscillators, the form of the equations will be very familiar and easy to understand and interpret.

I'm familiar only with the basic concept, none of the equations or means of analysing them. Last year I found some good animation that physically showed the difference of driven oscillators when they are in, and also out, of phase.
Calling this to mind helps me form an idea of why the speed of transmission would slow in a medium due to being "out of phase". If I picture each atom becoming the driver of the next atom in line, but as an out of resonance driver, I can imagine I understand where the slowing comes from.

What can you tell me about the atoms becoming dipoles? What happens in an atom to make it more positive or negative on one side than the other? (What comes to mind is the electron-orbitals being distorted somehow. Is that it?)

 

[Claude Bile]

What can you tell me about the atoms becoming dipoles? What happens in an atom to make it more positive or negative on one side than the other? (What comes to mind is the electron-orbitals being distorted somehow. Is that it?)

Atoms will become dipoles when an external electric field is applied. Essentially, the E field pulls positive charge in one direction and negative charges in the opposite direction (This will indeed distort the electron orbitals). The atom then posesses an intrinsic electric dipole whose electric field opposes the applied field, reducing the overall electric field.

Claude.

 

[zoobyshoe]

Thanks very much Claude. All that is extremely interesting, and I appreciate your taking the time.

 

[Claude Bile]

Not a problem, I'm glad you found it interesting.

Claude.

 

[John Q Public]

"Original" vs "new" photon

This subject is very old but, like many people who are trying to learn a new area of knowledge, I stumbled on it and found it fascinating enough to join the forum and comment. Hopefully someone knowledgeable can reply.

I see a debate went on between passing of the photon and absorption/readmission of a "new" photon. The latter puzzles me greatly when I try and envision this with glass, and the the primary reason is that the "new" radiation would be emitted isotropically (equally in all directions).

When I look through a window I see a fairly clear picture of what is on the other side, and it does not change in time due to the properties of the glass (unless something actually changes on the other side of the window. If photons are being absorbed and re-emitted isotropically then the orderly "information" about the picture we see looking through the glass would be quickly lost. That fact alone tells me with certainty that photons are NOT being absorbed and re-emitted. That process would be something more like diffusion/translucency, where energy is coming through but in an "unorderly" fashion.

Another issue would be partial absorption. I have read that a photon may lose part of it's energy exciting an electron, and then continue on to continue this process until depletion of all energy. If light passing through glass did this then a photon of "green" light would no longer be green. since frequency*wavelength = c and energy is directly proportional to the frequency, if the energy goes down the wavelength of the photon must increase to a lower energy color. In other words, unless almost all interactions were for the full photon energy of every photon, you would not see a clear picture in the right colors in the process were absorption and readmission.

I can see no way that light moves through glass via an absorption/readmission mechanism. It seems fairly clear that the original photons from the sun pass through the glass. It also seems that they actually are from the sun because if the original sun's photons were absorbed by anything on the way, the readmission would be isotropic and the chances of the re-emitted photon being in the original direction would be about nil. The sun's power would be highly diffused by the atmosphere. Would we be able to even see the sun and would shadows exist?

Can also see no way that the speed of light in glass is slowed by "wandering" of the photons around molecules, atoms, etc. Seems that all of the gazillions of photons that pass through the glass interact with the glass in the exact same way, which causes the entire image to be refracted in an orderly fashion.

A belated thank you to all who posted on this thread. Anyone know of good reading material out there? This subject is absolutely fascinating!

 

[ZapperZ]

Here's a very naive description of it, and I'm afraid, as much as I hate it, I have to use an analogy.

If you have a child on a swing, the BEST way for you to impart any kind of motion to the swing would be only on one particular direction. Any other direction would not be effective in causing a "proper" motion of the swing. Thus, if a group of "force carriers" were to come along in the correct direction, then they can impart a motion to this swing.

Now, imagine that there are many, many swings in a particular area. However, these swings are oriented in many different directions, even randomly oriented. This means that if I have "force carriers" coming in all different directions, most of them are bound to encounter a swing oriented along just the right direction to cause motion.

OK so far?

Now, if you have read our Physics Forum FAQ on this subject, you would have come across an explanation on why we have light transmission through solid. The problem now gets a bit complicated. If the force carriers that I've talked about above has a frequency at the same natural frequency as the swing, then that force is absorbed by the swing. This produces no retransmission of that force carrier because its energy has been absorbed by the swing system. In the phonon case, it means that there is a phonon mode of the solid and the photon has been absorbed as a lattice vibration (heat). This solid will be opaque to that light frequency.

However, if the "force carrier" isn't at the same frequency as the natural frequency of the swing, the swing cannot sustain such vibration and will simply transfer it to other swings or simply dumps it out. In a solid, this means that there is no phonon mode for that frequency of light and no absorption occurs via lattice vibration.

The fact here is that light has both energy and momentum. That momentum can only be transferred to a system that, to put it in simple form, is oriented in the right configuration for that momentum (or polarization). So this is not identical to an absorption by an atom, let's say, in which the photon's momentum has been taken up by the whole atom, and the reemission of another photon can be isotropic since momentum conservation can easily be taken up by that massive atom.

Zz.

 

[kcodon]

To understand what is happening when light passes through a medium, you need to understand what polarisation is. When one applies an electric field to an atom, the electric field causes the electron(s) to be slightly displaced from equilibrium, which forms an electric dipole. This pertubation can propagate through a medium, much the same way as if you jiggle a length of rope at one end, the pertubation (the jiggle), propagates down the length of the string. The energy and momentum of the incident light is carried in this wave, so essentially the light is, for all intents and purposes, traveling through the medium.

Why does light propagate slower in a medium than in a vacuum? As a perturbed atom passes the pertubation on to the nest atom, there is a slight shift in the phase of the oscillation of the two atoms. The culmination of these phase shifts manifests itself as a reduced velocity.

In a driven oscillator, the motion of the body undergoing oscillation is out of phase with the driving force when one is not on resonance. In the case of transmission through a medium, it is a necessary condition that the driving force not be on a resonance, otherwise absorption will result. Now consider the case where the motion of one electron (A) is driving the motion of the next (B). Electron B is out of phase with electron A, i.e. there is a net phase shift between A and B. Now consider a medium as a long path of electrons, the accumulated phase shift results in a slowed velocity.

What a nice explanation, and solves the isotropic worries about absorption and emission. However I do have some questions.

Firstly, when you talk of the "phase" of the electrons, I assume you mean the phase of their oscillating dipoles. Ah, and we you talk about electrons A and B (I'll stick with dipole A and B from now on, as involves all electrons, not just one I believe), being initially out of phase with each other, is this due to the fact that all the electrons in the atoms/molecules of a substance are continually moving, generating very quickly fluctuating dipoles? In chemistry this leads to the temporary dipole-dipole forces of attraction, if that helps anyone understand my attempt at an explanation. So anyway this means that for an atom A and B, their associated random electron distribution will result in the random dipole A and B, that are consequentially out of phase. Am I right so far?

Anyway, now the "phase shift". Firstly I don't have the correct terminology, so I'll just point out a few things before my explanation. Let's consider all the dipoles that are already oscillating in a nice sinusoidal pattern, similar to the wave on a string. Although they are not in phase with each other, they are in a "sinusoidal phase" so to speak; always constantly out of phase with the dipoles beside. I hope you understand what I'm trying to say; the reason will be apparent soon.

So continuing with the phase shift; is this simply the fact that once dipole A is set up by the photon, that dipole B must reorient itself (from its random temporary dipole), so as to become part of the "sinusoidal phase". This reorientation takes time, as the electrons can only travel at a finite speed, thus the phase shift, and light traveling at a reduce speed through transparent media.

Could anyone provide clarification as to whether I'm on the right path with that thinking?

However moving onward, one then considers an argument for minimising the phase shift; increasing the frequency. Surely at a high enough frequency of light, the dipole A would be oscillating so fast, that it would quickly move to dipole B's position, minimising phase shift (i.e. A moves to B, instead of B moving to A), and thus meaning higher frequency light would travel faster through a medium, which I don't believe occurs. One would think though that there was a maximum frequency at which the dipoles could oscillate, however the effect would still be noticeable. One might also argue that at such high frequency, it would result in heat loss pretty quickly, i.e. absorption.

I assume the isotropic problem is solved due to the orientation of the dipoles relative to the medium? I would have assumed dipole A would have induced dipoles in all the surrounding atoms, not only atoms B...thus the isotropic problem is back?

With absorption, I believe it is stated that resonance occurs so that the amplitude of the electrons in a dipole, is sufficient to transfer energy to the atoms (nuclei?), and thus heat and absorption. So really it is the amplitude of the movement of the electrons we are concerned about that causes absorption. Should this be affected by the varying frequencies of photons, I am unsure of. It would be affected by varying strength EM fields, but whether photons have varying strength EM fields, I am unsure again. Also, there must be more than one resonant frequency, by observations, and I believe this is due to standings waves along these phonons.

Also should anyone have any ideas on how this can lead to refraction, I would be greatly interested.

Thanks anyway, and sorry if its all a bit much, but might as well get it all out at once.

Kcodon

 

[Claude Bile]

Firstly, when you talk of the "phase" of the electrons, I assume you mean the phase of their oscillating dipoles. Ah, and we you talk about electrons A and B (I'll stick with dipole A and B from now on, as involves all electrons, not just one I believe), being initially out of phase with each other, is this due to the fact that all the electrons in the atoms/molecules of a substance are continually moving, generating very quickly fluctuating dipoles?

The atoms at equilibrium are assumed to have no net dipole. You can think of each atom in the chain as a spring - the applied E field results in a force that pulls the spring apart from equilibrium resulting in the formation of a dipole. Since the strength of the field determines how far the atom (or the electrons really since they are so much lighter than the nucleus) are displaced and thus how strong the dipole is. It follows that if the applied E field varies sinusoidally, then the dipole varies sinusoidally as well.

So continuing with the phase shift; is this simply the fact that once dipole A is set up by the photon, that dipole B must reorient itself (from its random temporary dipole), so as to become part of the "sinusoidal phase". This reorientation takes time, as the electrons can only travel at a finite speed, thus the phase shift, and light traveling at a reduce speed through transparent media.

One atom stretching to form a dipole results in a second E-field component that is due to the dipole. To a 1st approximation, we consider only the dipole contribution to the E-field when analysing the motion of the next atom in the chain. What is crucial here is that the driving force is NOT in phase with the motion of the dipole - because they are not being driven at a resonance. It is this phase difference that results in light slowing down in a medium.

Claude.

 

[kcodon]

The atoms at equilibrium are assumed to have no net dipole. You can think of each atom in the chain as a spring - the applied E field results in a force that pulls the spring apart from equilibrium resulting in the formation of a dipole. Since the strength of the field determines how far the atom (or the electrons really since they are so much lighter than the nucleus) are displaced and thus how strong the dipole is. It follows that if the applied E field varies sinusoidally, then the dipole varies sinusoidally as well.

I know this; what I was trying to understand was how there was a phase shift that could result in a slower propagation through the medium. It seems however that what I mentioned about how it takes time for a dipole to reorient itself is incorrect...?

One atom stretching to form a dipole results in a second E-field component that is due to the dipole. To a 1st approximation, we consider only the dipole contribution to the E-field when analysing the motion of the next atom in the chain. What is crucial here is that the driving force is NOT in phase with the motion of the dipole - because they are not being driven at a resonance. It is this phase difference that results in light slowing down in a medium.

Are you saying that the photon is not carried within the dipole to dipole movement? I.e. the dipoles flow in the sinusoidal EM wave form (created by the photon's EM field) and the EM field of the photon remains intact and travels through the medium independently? (I don't believe this is what you are saying, as it appears contradictory to other comments of yours). Would then the driving force be the EM field of the photon, which is out of phase with the EM field of the dipoles? However this still suggests the EM wave of the photon traveling at c, thus no slowing of wave...? How a phase difference between the two causes the slowing, I am still unsure.

I am now somewhat confused...

Kcodon

 

[John Q Public]

"I am now somewhat confused..."

Thanks for mentioning that, I suddenly feel better! :)

Still trying to get a handle on all of this. Does everyone agree that transmission through a medium is true transmission and not absorption/re-emission? If the consensus is the latter then I am totally lost here.

The refraction and slowing down at the medium surfaces is fascinating. There are some nice comments about the electric field interactions between the photon and the medium atoms propagating the EM wave and I can accept that at face value, at least for the moment. Since each material has a different refractive index and each material has a different molecular structure that supports the fact that the molecular structure is interacting with the photon/wave to change it's direction. How does the refraction occur at the surface? The photon/wave is changing direction the moment it enters the material, and keeps that direction until it exits, at which point it resumes it's original direction.

If the photon/wave is slowing down then is it safe to say it is imparting energy to the medium to set up the dipole in the medium atoms? Also, if the medium is amorphous (glass) then how in the heck does the photon/wave keep the same direction as it moves from molecule to molecule in the structure of the medium? It's not like there is a well laid out path of molecules to relay dipoles along the refractive path. By definition glass has no long range ordered molecular structure. This theory seems to have the photon going in whatever direction the molecular structure takes it, yet all the photons passing through the material take the same path relative to their entry point (preserving the image we see looking through the glass).

Lastly I am wondering about the e field interaction. If ANY energy is lost during the interaction then the photon cannot come out of the material with the same energy it entered. Is it possible for all these interactions to be perfectly lossless? How would the loss of energy manifest itself? It seems the frequency of the exiting photon would have to come down.

 

[kcodon]

Also, if the medium is amorphous (glass) then how in the heck does the photon/wave keep the same direction as it moves from molecule to molecule in the structure of the medium? It's not like there is a well laid out path of molecules to relay dipoles along the refractive path. By definition glass has no long range ordered molecular structure. This theory seems to have the photon going in whatever direction the molecular structure takes it, yet all the photons passing through the material take the same path relative to their entry point (preserving the image we see looking through the glass).

Excellent question...and I have no idea as to the answer, but I hope someone could provide one.

It was similar to what I was getting at when I said

I would have assumed dipole A would have induced dipoles in all the surrounding atoms, not only atom B...thus the isotropic problem is back?

For even with a crystallic structure I would think a dipole would affect and induce dipoles in all atoms around it... 3D...not neccesarily just those "in front" of it. Possibly the direction in which the dipole is relative to the glass may solve this problem.

Kcodon

 

[Claude Bile]

Are you saying that the photon is not carried within the dipole to dipole movement? I.e. the dipoles flow in the sinusoidal EM wave form (created by the photon's EM field) and the EM field of the photon remains intact and travels through the medium independently? (I don't believe this is what you are saying, as it appears contradictory to other comments of yours).

Back to the string analogy, think of the photon as the hand that waves the string at one end, causing a wave to propagate down the string. You touch on the question of whether the propagating EM field is due to the dipoles or the original photon. As it turns out the question is somewhat moot - what matters is the presence of an EM field oscillating at some frequency.

Would then the driving force be the EM field of the photon, which is out of phase with the EM field of the dipoles?

The phase difference is cumulative remember, as the wave propagates through more atoms. As before, whether the wave is a "photon" or a "dipole" is irrelevant, what is important is that there is an EM wave present.

How a phase difference between the two causes the slowing, I am still unsure.

Wave velocity is equal to \frac{\omega}{k}. \omega is equal to no of radians subtended per second. By adding a negative phase at regular intervals, \omega is reduced since because the radians subtended per second (on average) is now less, thus the wave slows down.

Sorry for the confusion, hopefully that clears some things up.

Claude.

 

[Claude Bile]

Still trying to get a handle on all of this. Does everyone agree that transmission through a medium is true transmission and not absorption/re-emission? If the consensus is the latter then I am totally lost here.

It is true transmission.

The refraction and slowing down at the medium surfaces is fascinating. There are some nice comments about the electric field interactions between the photon and the medium atoms propagating the EM wave and I can accept that at face value, at least for the moment. Since each material has a different refractive index and each material has a different molecular structure that supports the fact that the molecular structure is interacting with the photon/wave to change it's direction. How does the refraction occur at the surface? The photon/wave is changing direction the moment it enters the material, and keeps that direction until it exits, at which point it resumes it's original direction.

I think we are straying into thinking of a photon as a particle here rather than a wave. The particle nature of the photon doesn't really enter the picture in the case of transmission, since there is no localised interaction as there is in the case of absorption and re-emission. Refraction occurs purely because part of the wavefront slows down, to maintain continuity at the interface, the wave must change direction as it crosses the interface.

If the photon/wave is slowing down then is it safe to say it is imparting energy to the medium to set up the dipole in the medium atoms?

The wave does impart energy to the medium, but it gets it all back in the end (i.e. the interaction is elastic). The imparting of energy has nothing to do with the speed of the wave however.

Also, if the medium is amorphous (glass) then how in the heck does the photon/wave keep the same direction as it moves from molecule to molecule in the structure of the medium? It's not like there is a well laid out path of molecules to relay dipoles along the refractive path. By definition glass has no long range ordered molecular structure. This theory seems to have the photon going in whatever direction the molecular structure takes it, yet all the photons passing through the material take the same path relative to their entry point (preserving the image we see looking through the glass).

The key to understanding photon transmission is that is an entirely non-resonant process - there is no localised interaction between photons and atoms, so we can explain everything using waves. Since electronic wavefunctions fill the entire medium, what we are essentially doing is expressing photon transmission as a "ripple" in the electronic wavefunctions that permeate through the medium. There is no need for any precise arrangement of the atoms and molecules themselves.

Again, reiterating my earlier points, there is no point making the distinction between a "photon" ripple and a "dipole" ripple. The important thing is that there is a ripple.

Lastly I am wondering about the e field interaction. If ANY energy is lost during the interaction then the photon cannot come out of the material with the same energy it entered. Is it possible for all these interactions to be perfectly lossless? How would the loss of energy manifest itself? It seems the frequency of the exiting photon would have to come down.

All energy imparted to the medium is temporary in the case of a lossless medium. Real media will have losses because, in a real atom, there is still a probability (sometimes significant) that a photon WILL undergo a localised interaction, be it an absorption event or an elastic scattering event. These losses however is a reduction in the number of photons, not the energy of each individual photon.

Some interactions, such as Raman scattering will result in photons losing energy, and thus reducing their frequency.

I apologise if my recent responses seem a bit incoherent compared to my earlier ones, but this was an old thread recently "resurrected" as it were and my brain has had to start over .

Claude.

 

[Nacho]

Good topic. The subject of light reflection and diffraction facinates me.

This is what Richard Feynman says about it in his book "QED" (a grouping of lectures for a nontechnical audience). This passage isn't specifically about diffraction; it's about partial reflection of light through glass. Can diffraction be considered a complement to reflection?

Page103 & 104:

Step #1: A photon is emitted by the source at a certain time.

Step #2: The photon goes from the source to one of the points in the glass.

Step #3: The photon is scattered by an electron at that point.

Step #4: A new photon makes its way up to the detector.


For step #4, the detector is on the same side of the glass as the emitter, detecting the photons reflected off of the glass. Also notice his using the term "new photon". He further explains the "step 3", that the photon can scatter in any direction, but if you add up the amplitude for the scatter in all those directions, they all cancel out except in the direction that the light actually emerges .. I guess bringing order to a random event?

I have a question about relection and/or diffraction that I hope can be answered:

Does anything **unexpected** happen when the intensity of the light is varied, or the density of the reflecting/refracting material is varied? By **unexpected**, I mean does doubling the intensity of the light always double the amount of light that is reflected (up to a point .. without melting the reflecting/refracting medium)? Does altering the density change the refractive index? I guess what I'm getting at has to do with the number and placement of electrons in the medium available to scatter a photon. As the intensity of light gets more and more, or the refracting medium's density gets lessor and lessor, is there be some point where there are not enough electrons available to absorb the photons, and so would there be 2 paths of exit from the medium: 1 a refracted path and 1 a straight-through path?

 

[GT1]

If there is no interaction between Photons and atoms, how can the speed of light be slower then c (speed of light in vacuum) ? what slow down the photons ? (or the wave)
don't you need external energy the slowdown something ?
and how can the speed of light in some mediums be faster then c - what makes the photons (or the wave) go faster ?

 

[kcodon]

Back to the string analogy, think of the photon as the hand that waves the string at one end, causing a wave to propagate down the string. You touch on the question of whether the propagating EM field is due to the dipoles or the original photon. As it turns out the question is somewhat moot - what matters is the presence of an EM field oscillating at some frequency.

The phase difference is cumulative remember, as the wave propagates through more atoms. As before, whether the wave is a "photon" or a "dipole" is irrelevant, what is important is that there is an EM wave present.

I am in agreement with this...what I said before was an attempt to understand the phase difference.

Wave velocity is equal to \frac{\omega}{k}. \omega is equal to no of radians subtended per second. By adding a negative phase at regular intervals, \omega is reduced since because the radians subtended per second (on average) is now less, thus the wave slows down.

I know what you mean about how the phase difference will be cumulative, and how it will slow down, however I do not see where this "phase difference" comes from...I do not believe you have elaborated on this so far...I would much appreciate if you could explain this phase difference in some physical manner.

Here is what I think (from my previous post):

So continuing with the phase shift; is this simply the fact that once dipole A is set up by the photon, that dipole B must reorient itself (from its random temporary dipole), so as to become part of the "sinusoidal phase". This reorientation takes time, as the electrons can only travel at a finite speed, thus the phase shift, and light traveling at a reduce speed through transparent media.

Could you verify whether this is correct, or not so?

Thanks,

Kcodon

 

[M Grandin]

Huygens Principle

This leads to interesting thoughts about "Huygens Principle", according to for instance site

http://www.mathpages.com/home/kmath242/kmath242.htm

As hinted there, also traveling matter may be regarded as fluctuating waves between -C and + C velocities, where resulting "group velocity" is the observed velocity. Appear fruitful trying generalize Huygens principle, that could perhaps also explain "Einstein relativity" in a more common-sense way.

 

[Claude Bile]

I know what you mean about how the phase difference will be cumulative, and how it will slow down, however I do not see where this "phase difference" comes from...I do not believe you have elaborated on this so far...I would much appreciate if you could explain this phase difference in some physical manner.

The phase difference is a standard property of any driven oscillator. If you drive an oscillator with a resonance at frequency \omega_0 at some non-resonant frequency \omega, then the driving force will not be in phase with the motion of the oscillator.

The converse case, where the driving force IS in phase with the motion of the oscillator is the very definition of a resonance, since the driving force is ALWAYS in the same direction as the oscillator motion, hence energy is coupled into the oscillator very efficiently. Since we are not at resonance in the case of transmission, there must be some phase difference between the driving force (i.e. the incident E field) and the motion of the oscillator (i.e the response field).

Claude.

 

[kcodon]

Thanks Claude.

I haven't been able to reply as soon as I would have liked, however I'd like to thank you for your perseverance! I finally found a site that explained the phase difference you keep referring to, and I think I now see how this applies to the transmission of light. Its gloriously more simple than things I was imagining, so thanks again,

Kcodon

 

[GT1]

Thanks Claude.

I finally found a site that explained the phase difference you keep referring to, and I think I now see how this applies to the transmission of light. Its gloriously more simple than things I was imagining, so thanks again,

Kcodon

Can you please add the link?

 

[kcodon]

The phase difference is a standard property of any driven oscillator. If you drive an oscillator with a resonance at frequency \omega_0 at some non-resonant frequency \omega, then the driving force will not be in phase with the motion of the oscillator.

The converse case, where the driving force IS in phase with the motion of the oscillator is the very definition of a resonance, since the driving force is ALWAYS in the same direction as the oscillator motion, hence energy is coupled into the oscillator very efficiently. Since we are not at resonance in the case of transmission, there must be some phase difference between the driving force (i.e. the incident E field) and the motion of the oscillator (i.e the response field).

Claude.

Hi GT1,

Claude puts it nicely, but this site http://perlnet.umaine.edu/IMT/FHM/IG%20FHM.pdf"
helped me understand a bit more. I doubt you will find it useful, as it's actually like a lesson plan and not exactly an orthodox explanation, however see for yourself.

I never understood how two dipoles could have this phase difference. However basic idea - at least I think - is that the dipole A drives dipole B when dipole B is at max velocity, opposed to at max displacement, if that makes sense. Think of a swing...one would intuitively expect that the dipole would be similar to the swing case...providing the driving force (pushing swing) at max displacement (when swing at peak), however I don't think this is the case...the drive is provided not at the peak. This is the phase difference. How this slows down light transmission: think of a line of swings. If each one doesn't drive the other at the "peak", then can you imagine how that slows it down?

Another site to see the effect of this phase difference is http://webphysics.davidson.edu/faculty/thg/physlets/html/driven_sho.html"
Hope it helps,

Kcodon

 

[Claude Bile]

Thanks Claude.

I haven't been able to reply as soon as I would have liked, however I'd like to thank you for your perseverance! I finally found a site that explained the phase difference you keep referring to, and I think I now see how this applies to the transmission of light. Its gloriously more simple than things I was imagining, so thanks again,

Kcodon

No problem, glad to see I was a help rather than a hindrance .

Claude.

 

[GT1]

Thanks Kcodon!

 

[catkin]

Why does light propagate slower in a medium than in a vacuum? As a perturbed atom passes the pertubation on to the nest atom, there is a slight shift in the phase of the oscillation of the two atoms. The culmination of these phase shifts manifests itself as a reduced velocity.

Thanks for that; it's the closest I've found to a satisfying explanation but I'm still curious.

How wrong is the following wild theorising? The photon is a probability density wave function, not actually manifested at any place along that wave function until it interacts with another probability density wave function when both probability density wave functions have to collapse. In the case of light (photon) transmission through a medium, the photon's wave continues to move at the speed of light but the centre of probability density (the place the photon is most likely to manifest if it must) slides backwards along the wave. The only way we can measure the photon is by making it manifest thus all our observations relate to the the centre of probability density and so indicate that the photon is traveling slower in the medium than in a vacuum.

I'm particularly worried by "The only way we can measure the photon is by making it manifest"; what about all the experiments that demonstrate wave-like properties? And by the implicit assertion that the centre of probability density is made to slide backwards without any interaction that would require collapse of the wave function.