How Does Light Behave When It Passes Through Different Mediums?

[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.