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...
Quantum Electrodynamics does indeed explain. Feynman wrote a book (with help because he hated the effort required to write books) s entitled QED to cover the topic. My guess is that there still is a series of lectures given at the University of Auckland on-line. Googke for it.
Bill
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An old man would be better off never having been born.
Not just the relatively new wave theory. "Snell's" law and ray theory go back to Alhazen.
Not really *that* simple. The presence of electrons' charges absorbs some energy from passing photons and slows them down; the charges then oscillate out of phase with the photons, generating their own alternating field which interferes with the incoming photons. The resultant of the interference, with the same frequency but shorter wavelength as the incoming photons, then is either transmitted into the material containing the electrons, or reflected out of it, depending on the angle of incidence.
Transmission, reflection, and refraction at a surface are governed by the ratios of the refractive index on both sides of the surface.
The whole atom does not reflect photons, the electrons do, and orbitals are distorted by being involved in bonds that hold atoms together. An atom's electrons on the surface of an object see half of the universe as "constricted" by the fields of other atom's electrons, the other half as "free" of that constriction. The photons' energy can't be held onto by the electrons forever, they _have_ to let go of it, and the most likely direction is that with the lower index.
RichD wrote in news:7d8aade9-635f-4ef9-9c04- snipped-for-privacy@o13g2000vbl.googlegroups.com:
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.
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.
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.
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.)
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.
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).
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.
The particle of light has to be in either the electric or magnetic wave. Glass does not expand when light passes through. This is the evidence that it does not absorb. Transparency involves not absorbing. The field of the atom slows light in a medium.
This is not how reflection works, in the QM picture. The energy in the photons are not absorbed. The electrons don't gain energy to go to higher absorption levels. Your model is actually how physicist describe photoluminescence. Photoluminescence acts exactly the way you just described. Photoluminescence occurs in all directions. There is no law in photoluminescence that "the angle of incidence equals the angle of reflection."
Actually, the question does come up in the wave picture of reflection. The answer here is that the atoms are more closely spaced that the wavelength of the light. So there is a strong diffraction of spherical light waves scattered from nearby atoms. The "high intensity" band caused by the diffraction of light happens to be in the direction determined by the law of reflection. This does not answer your question. However, your queHowever, your question is whether one can force
Reflection does not have a time delay. Reflection occurs "instantaneous." The atom does not have time to "forget" the original direction. At least not in reflection. Again, you are thinking of photoluminescence. Photoluminescence does occur after a time delay. The emitted photon in photoluminescence does not remember where the original photon came from. The emitted photon is fired in a direction which is uncorrelated with the original direction.
See my other post.
Yes it is. However, I don't think we have to go there. I will address your questions again in a different post. In this post, I just want to point out some errors embedded in your model. Your error is that you read something about how quantum mechanics applies to photoluminescence. The real issue is how quantum mechanics applies to reflectivity.
Einstein questioned his photon and said he could never reconcile it with the wave. He questioned what he won the Nobel Prize for. What wave is the particle of light in? the electric opr magnetic wave? Mitch Raemsch
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.
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.
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. =A0 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.
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
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 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.
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?
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