0 Members and 1 Guest are viewing this topic.
Couldn't it be that the light wave is only losing amplitude, and not frequency?
A 'photon' does not have a frequency. You can if you like see it as equivalent to a wave, assuming a lot of 'photons' here, but a photon express itself as 'particle', not as a wave.
Yes, if we're speaking photons they lose 'energy/momentum'. I've seen a lot of new formalism jumping between 'photons' and waves the last decades, without changing the original definitions? wonder why, if now a photon would have both a frequency and a wavelength?
Back to the original post: yes, the light redshifts. […] If the photon were to reflect from the mirror with the same frequency at which it arrived, then the photon wouldn't lose any energy. But the mirror is now moving, so the mirror has gained some energy. Since total energy has to be conserved, this can't happen, and indeed the photon must lose a bit of energy in reflection so it redshifts.
The simplest physical explanation I've seen is if you imagine you fire a photon at a mirror that's at rest. Once the photon hits the mirror and reflects, the mirror is moving away from you, so the reflected photon is now being emitted by a mirror that is moving away from you which means they experience a Doppler shift towards lower frequencies.
So, if you think of your bathroom mirror, essentially fixed, then you should not get any redshift/blueshift.
When it comes to a 'wave' crossing a border/density, as in glass, it will be diverted. That's referred to as 'refraction' and it will change the waves wavelength, but not its frequency. A frequency is defined as the 'number of cycles per unit time' that you measure whereas a wavelength is defined as the distance between one peak, or crest, of that same wave of light. The frequency is what defines the color we see, so if we assume that the frequency change with a wave getting reflected then all mirrors should change the color of what they reflect as the light travels through glass first to then interact with some silvery medium before getting reflected. Only in space, assuming a perfect elastic collision, could you expect the color to stay the same for a beam of light.The index of refraction is 'n' ( n = c / v ) where 'c' is lights speed in a vacuum, 'v' is your new speed inside that medium, or density. And to get to the density inside something?The index of refraction can also be stated in terms of wavelength (sorry, tried to upload the equation but? The upload won't work?) Ah well ...sorry, you cannot view external links. To see them, please
REGISTER or LOGIN