Timeline for Why does the light travel slower in denser medium?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 24, 2019 at 4:05 | comment | added | Ján Lalinský | In classical EM theory, EM field obeys linear equations, so net field is just a sum of primary and secondary field. The primary field is unchanged by the secondary field, they both contribute to total EM field. If primary field is plane wave, then it continues on its path, undisturbed by the charge or its field. The only thing the charge changes it produced additional field component. | |
| Nov 23, 2019 at 22:42 | comment | added | Antonios Sarikas | @JánLalinský Thanks for the answer. But I still don't get it. Lets consider an em wave interacting with a single charge in space. What bothers me is if we treat light as photons then the total number of photons that hit the charge must be the same that come out after interact with it (conservation of energy). But if we treat light as an em wave then as the (electric) field oscillates when it meets the charge the latter start to oscillate so it emitts radiation. But does the initial em wave continue to propagate like a wave in a lake (e.g. when it meets a ball in the surface of the lake) ? | |
| Nov 20, 2019 at 21:48 | history | edited | Ján Lalinský | CC BY-SA 4.0 |
added 277 characters in body
|
| Nov 20, 2019 at 21:16 | comment | added | Ján Lalinský | @adosar EM energy can be attributed either to a region of space (as often done with Poynting's expressions), or to some subset (at least two) of charged particles and some region of space (as in Tetrode/Fokker/Frenkel/Feynman-Wheeler theories of point particles), but it cannot be, in general, assigned to some given component of EM field that obeys Maxwell's equations (which secondary EM wave is). Whatever the secondary wave looks like, EM energy + matter energy is locally conserved. | |
| Nov 20, 2019 at 21:14 | comment | added | Ján Lalinský | @ado sar That is a good question. We know that secondary radiation exists (from experience) and we believe energy is locally conserved (an extrapolation of definition of energy allowed by experience). Sometimes it seems as if presence of secondary wave near accelerating charged object means there is some additional EM energy that appeared out of nowhere (not from the primary wave). But this is just confusion about the concept of EM energy and how it can be assigned to regions or particles. | |
| Nov 20, 2019 at 20:53 | comment | added | Antonios Sarikas | @JánLalinský But how is it possible that secondary waves to exist? Doesn't this violate the conservation of energy? | |
| Aug 14, 2017 at 23:07 | comment | added | Ján Lalinský | I am talking about an EM wave in material medium, which is a model of a light wave, but the important part in this explanation is relation of electric field to current density, not energy of the wave or the concept of a photon. | |
| Aug 14, 2017 at 11:18 | comment | added | PDiracDelta | I think I've found a good explanation here: physics.stackexchange.com/a/476/51901 (this question now apparently has been marked as a duplicate) | |
| Aug 14, 2017 at 11:03 | comment | added | PDiracDelta | I'm trying to understand what you are saying, but if not light, then what kind of wave are you talking about? AFAIK a light and EM wave are the same thing. Also, to me your claim that the frequency remains the same is non-trivial. | |
| Aug 12, 2017 at 23:27 | history | edited | Ján Lalinský | CC BY-SA 3.0 |
added 193 characters in body
|
| Aug 12, 2017 at 23:22 | comment | added | Ján Lalinský | @PDiracDelta, the question is about index of refraction. Index of refraction is ratio of phase velocities of a wave in two media. Since frequency is the same in both media (linear media) and since $v = f \lambda$, decrease in wavelength implies decrease in phase velocity $v$. The energy of a photon or light wave has nothing to do with this explanation. | |
| Aug 11, 2017 at 22:38 | comment | added | PDiracDelta | This explanation cannot be correct. You state that "the resulting wave in the medium will have shorter wavelength hence lower velocity" but this is incorrect. If you change the wavelength (or the frequency, because frequency = c / wavelength) you only change the energy of the photon / light wave (and thus colour) but not the speed, which is fixed as long as it remains in the same type of medium. Indeed though as @arax notes, this fixed speed actually depends on the type of medium and can become smaller than c, but I do not know the correct explanation. | |
| S Jul 22, 2017 at 20:41 | history | suggested | Master Drifter | CC BY-SA 3.0 |
added several grammatical articles where necessary
|
| Jul 22, 2017 at 18:01 | review | Suggested edits | |||
| S Jul 22, 2017 at 20:41 | |||||
| Apr 24, 2016 at 17:55 | review | Suggested edits | |||
| Apr 24, 2016 at 18:08 | |||||
| Mar 29, 2014 at 8:11 | history | edited | Ján Lalinský | CC BY-SA 3.0 |
added 140 characters in body
|
| Mar 29, 2014 at 8:05 | history | answered | Ján Lalinský | CC BY-SA 3.0 |