photons and reflection
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From: Androcles on 28 Nov 2009 00:55 "RichD" <r_delaney2001(a)yahoo.com> wrote in message news:7d8aade9-635f-4ef9-9c04-ebafbc5b7a42(a)o13g2000vbl.googlegroups.com... > According to the wave theory of light, angle of > incidence equals angle of reflection. No problem, > in theory or fact. > > However, per QM, light falls as a 'rain' of photons. > What happens then? According to the Newtonian theory of bouncing balls, angle of incidence equals angle of reflection. No problem, in theory or fact. According to the quantum theory of light, angle of incidence equals angle of reflection. No problem, in theory or fact. > As I understand it (big qualifier > there), the photons are absorbed by surface atoms. Not true. Only photons of the correct wavelength are absorbed, the rest bounce away. A white surface reflects them, a black surface absorbs them, a green surface reflects green photons and absorbs red and blue ones.
From: John Devereux on 28 Nov 2009 02:31 Skywise <into(a)oblivion.nothing.com> writes: > RichD <r_delaney2001(a)yahoo.com> wrote in news:7d8aade9-635f-4ef9-9c04- > ebafbc5b7a42(a)o13g2000vbl.googlegroups.com: > >> I never studied quantum field theory, maybe it's >> explained there... > > I will add my voice to the recommendation of QED by Feynman. It > was a very enjoyable read, and you don't need to be a math major > to comprehend it, as there is no math. > > Brian For those searching, the full title is actually "QED: The Strange Theory of Light and Matter" e.g. <http://www.amazon.com/QED-Strange-Princeton-Science-Library/dp/0691125759> An amazing book. -- John Devereux
From: George Herold on 28 Nov 2009 09:31 On Nov 27, 7:24 pm, RichD <r_delaney2...(a)yahoo.com> wrote: > According to the wave theory of light, angle of > incidence equals angle of reflection. No problem, > in theory or fact. > > However, per QM, light falls as a 'rain' of photons. > What happens then? As I understand it (big qualifier > there), the photons are absorbed by surface atoms. > Electrons jump to higher energy orbitals, then fall > back to ground state, emitting photon(s) of its > characteristic spectrum. Simple.... > > This raises several questions, regarding geometry... > the aforementioned angles are defined relative > to a surface normal. But the surface is not truly > continuous, it's atomic and chunky. How does an > atom know where the 'normal' is? How does it > know which direction to fire its photons, after a > time delay? Does it have some sort of 'light > momentum' memory? > > I never studied quantum field theory, maybe it's > explained there... > > -- > Rich Rich, You need to separate the case of individual atoms, and (crystaline) solids. In solids the atomic states tend to form bands of energy states. Metals have 1/2 filled bands and give reflections. I don't think it is correct to think that only the surface atoms (electrons) are doing the reflection. The wavelength of visible light is much bigger than atomic spacing. Lots of electrons are taking part in the light-matter interaction. I'm not sure QED is going to help much. It's a long ways from QED to Solid state physics. Does QED even help do atomic physics? (I've never done any QED calculations.) George H.
From: PD on 28 Nov 2009 11:15 On Nov 27, 6:24 pm, RichD <r_delaney2...(a)yahoo.com> wrote: > According to the wave theory of light, angle of > incidence equals angle of reflection. No problem, > in theory or fact. > > However, per QM, light falls as a 'rain' of photons. > What happens then? As I understand it (big qualifier > there), the photons are absorbed by surface atoms. > Electrons jump to higher energy orbitals, then fall > back to ground state, emitting photon(s) of its > characteristic spectrum. Simple.... But photons also have *phase*, and this turns out to be crucial. You will get a better appreciation of how this influences things if you read a thin book by Fenyman called QED. > > This raises several questions, regarding geometry... > the aforementioned angles are defined relative > to a surface normal. But the surface is not truly > continuous, it's atomic and chunky. How does an > atom know where the 'normal' is? How does it > know which direction to fire its photons, after a > time delay? Does it have some sort of 'light > momentum' memory? > > I never studied quantum field theory, maybe it's > explained there... > > -- > Rich
From: whit3rd on 29 Nov 2009 14:29
On Nov 27, 4:24 pm, RichD <r_delaney2...(a)yahoo.com> wrote: > According to the wave theory of light, angle of > incidence equals angle of reflection. No problem, > in theory or fact. > > However, per QM, light falls as a 'rain' of photons. > What happens then? As I understand it (big qualifier > there), the photons are absorbed by surface atoms. No, that's not a useful way to proceed; you'd have to come up with a way to conserve momentum AND energy, but there aren't any one-photon/one-atom solutions that do it. Well, usually aren't (Mossbauer effect- look it up). > I never studied quantum field theory, maybe it's > explained there... The only useful part of quantum theory is that the light has a wave equation. The electrically-conductive flat surface of a mirror has a known internal electric field (zero) and that means that an incoming photon's electric wave has a bordering plane of null electric field, which only works if there's a second wave and a standing-wave kind of nullification occurs. The second wave, the reflection, plus the incoming wave has the right boundary condition at the mirror plane. The full quantum electrodynamics theory is both relativity and quantum in one piece, includes lots of stuff that isn't relevant to mirror reflection of light; a mirror is just like an antenna, it receives and/or transmits according to internal currents and all you really need to understand a mirror is a few of Maxwell's equations. Regular visible light is VERY LARGE particles, much bigger than an atom; the mirror surface isn't rough enough to matter to the extended electromagnetic wave that is the image of your razor during the morning shave. |