photons and reflection
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From: Rich Grise on 30 Nov 2009 14:59 On Fri, 27 Nov 2009 16:24:07 -0800, RichD 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... Look at a piece of aluminum foil. One side is mirror-smooth, such that you could see your reflection, if you could make it flat enough. The other side is matte, and doesn't give a mirror-like reflection. Does help at all? Cheers! Rich
From: BURT on 30 Nov 2009 16:50 On Nov 30, 11:59 am, Rich Grise <richgr...(a)example.net> wrote: > On Fri, 27 Nov 2009 16:24:07 -0800, RichD 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... > > Look at a piece of aluminum foil. One side is mirror-smooth, such that > you could see your reflection, if you could make it flat enough. The other > side is matte, and doesn't give a mirror-like reflection. Does help at all? > > Cheers! > Rich- Hide quoted text - > > - Show quoted text - Light comes from every angle but if energy is quantized their can be no rainbow or the full range of a prism spectrum. Mitch Raemsch
From: Salmon Egg on 30 Nov 2009 21:25 In article <pan.2009.11.30.19.59.55.233194(a)example.net>, Rich Grise <richgrise(a)example.net> wrote: > Look at a piece of aluminum foil. One side is mirror-smooth, such that > you could see your reflection, if you could make it flat enough. The other > side is matte, and doesn't give a mirror-like reflection. Does help at all? Use X-band sensitive eyes! Bill -- An old man would be better off never having been born.
From: BURT on 30 Nov 2009 21:33 On Nov 30, 6:25 pm, Salmon Egg <Salmon...(a)sbcglobal.net> wrote: > In article <pan.2009.11.30.19.59.55.233...(a)example.net>, > Rich Grise <richgr...(a)example.net> wrote: > > > Look at a piece of aluminum foil. One side is mirror-smooth, such that > > you could see your reflection, if you could make it flat enough. The other > > side is matte, and doesn't give a mirror-like reflection. Does help at all? > > Use X-band sensitive eyes! > > Bill > > -- > An old man would be better off never having been born. There is no way light can be quantized in energy comming out of the atom is it produces a full spectrum of energy levels. Mitch Raemsch - Still in the aether of time
From: George Herold on 1 Dec 2009 10:14
On Nov 29, 5:12 pm, BURT <macromi...(a)yahoo.com> wrote: > On Nov 29, 2:02 pm, Darwin123 <drosen0...(a)yahoo.com> wrote: > > > > > > > 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. > > > This statement isn't completely accurate. According to the wave > > theory of light, diffraction can greatly affect the angle of > > reflection. Gratings are made with periodic lines engraved on them. > > These periodic lines can greatly affect the reflection from the > > surface. > > Rough surfaces can spread out the angle of reflection of light by > > at least two ways. First, the microscopic facets on the surface can > > make the real angle of incidence different from the apparent angle of > > incidence. However, diffraction can also make the light spread out > > even more. > > > > However, per QM, light falls as a 'rain' of photons. > > > What happens then? > > > This question as stated is unclear to me. Quantum mechanics > > doesn't make the statement that light falls as a "rain of photons." > > Quantum mechanics is based on the duality of particles and waves. > > I think you meant something slightly different. Allow me to > > restate your question in terms of what I think you meant. Then I can > > answer this restated question. > > I think you were asking how the law of reflection could be > > explained that if one assumes that light is completely made of > > particles, with no waves. This type of model was used by Issaac > > Newton, hundreds of years before quantum mechanics. This "corpuscular" > > theory of Newton's is sometimes used as an approximation of quantum > > mechanics. It isn't very accurate except under very specific > > conditions. However, I think your question can be broken up into two > > questions. > > 1) Given the Newtonian picture of light as consisting of a rain of > > corpuscles, can one explain reflectivity? > > 2) How closely do Newton's corpuscles in Newton's theory of light > > resemble the photons in quantum mechanics? > > I think a little review of Newton's theory may be helpful > > here. > > In Newton's corpuscular theory, the corpuscle bounces off the > > surface due to forces between the surface and the corpusule. Three > > assumptions are made in this theory to get the law of reflection. > > A) The surface is extremely slippery, with no friction. Hence, the > > component of momentum parallel to the surface is completely conserved. > > B) The surface pushes back by conservative forces. Hence, the total > > mechanical energy of the corpuscle is completely conserved during the > > reflection. > > C) The surface is completely flat. > > The law of reflection is ttrue in the inertial frame of the > > surface if conditions A, B, and C are true. Thus, Newton's corpuscles > > precisely explain the reflection of light off a smooth surface. > > Now that we understand Newton's theory, which I believe was in > > the back of your mind, I can address your questions.> 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? > > > Condition C is no longer valid. Conditions A and B are still > > valid. Each atom is slippery and elastic. However, real surfaces are > > rough. > > Newton did not know as much as we do about atoms. So let me > > modify Newton's corpuscle theory just a little bit. > > Assume that the corpuscle of light is an extremely large sphere. > > In fact, let us assume better. The corpuscle of light has the shape of > > a sphere with a diameter equal to the wavelength of the "nonexistent" > > light wave. The "nonexistent" light wave has a wavelength several > > thousand times the diameter of the atom.>How does it > > > know which direction to fire its photons, after a > > > time delay? > > > The corpuscle is so big, it experiences the surface as being > > smooth. That is, the surface of the corpuscle touches several atoms > > when it collides with the surface. Condition>Does it have some sort of 'light > > > momentum' memory? > > > In Newton's corpuscle theory, the only memory is that of the > > conservation laws. As I stated, it is the conservation laws that cause > > the law of reflection to be valid. The basis of the conservation laws > > are the forces that the atoms exert. In this sense, the atom has a > > memory. By virtue of being elastic, the atom "remembers" the > > mechanical energy. By virtue of its being slippery, the atom > > "remembers" one component of momentum. > > > > I never studied quantum field theory, maybe it's > > > explained there... > > > I statement is a little bit like the question, "Is Newton's > > theory of light corpuscles ever a satisfactory approximation of > > quantum mechanics?" > > Now here is where Al gets to call me an "idiot". I am going to > > give a very qualified, "Yes." > > Conservation of energy and conservation of momentum are > > embedded in quantum field theory, just as they are embedded in quantum > > field theory (QED). In order to get the conservation laws, one has to > > apply symmetry conditions to Hamiltonians. According to Noether's > > theorem, every conservation law has to be associated with a > > corresponding symmetry property. This applies to Newtonian mechanics, > > classical relativistic mechanics, and to QED. > > In the abstract, the law of reflection is a result of two > > conservation laws. The wave-corpuscle-photon has to conserve both > > mechanical energy and the component of linear momentum that is > > parallel to the surface. Thus, any picture for the reflection process > > that includes these two conservation laws is a satisfactory model for > > the reflection process. This means that any picture that includes the > > corresponding symmetry conditions will also be a satisfactory > > picture. > > The picture of a large puffy corpuscle bouncing off these tiny > > atoms is a good phenomenological model since we can build the > > symmetries into the shape of the corpuscle. One can force fit just > > about any result by choosing the shape of the corpuscle. One may get a > > corpuscle to behave like a photon under a very narrow range of > > conditions. For what ever that is worth. > > No. There is no particle of light. It is easily demostratable as a > question that cannot be answered. > > Mitch Raemsch- Hide quoted text - > > - Show quoted text - "> No. There is no particle of light. It is easily demostratable as a > question that cannot be answered." What? You haven't heard of a PMT? (Photomultiplier tube) or the photoelectric effect? George H. |