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From: BURT on 29 Nov 2009 16:05
On Nov 29, 11:29 am, whit3rd <whit...(a)gmail.com> wrote:
> 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.

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.

Mitch Raemsch
From: Darwin123 on 29 Nov 2009 16:22
On Nov 27, 7:24 pm, RichD <r_delaney2...(a)yahoo.com> wrote:
>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.  
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."
>
> 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?
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

>How does it
> know which direction to fire its photons, after a
> time delay?  
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.
>Does it have some sort of 'light
> momentum' memory?
See my other post.
>
> I never studied quantum field theory, maybe it's
> explained there...
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.
From: BURT on 29 Nov 2009 16:59
On Nov 29, 1:22 pm, Darwin123 <drosen0...(a)yahoo.com> wrote:
> On Nov 27, 7:24 pm, RichD <r_delaney2...(a)yahoo.com> wrote:>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.  
>
>       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."
>
> > 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?
>
>          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
>
> >How does it
> > know which direction to fire its photons, after a
> > time delay?  
>
>      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.>Does it have some sort of 'light
> > momentum' memory?
>
>      See my other post.
>
> > I never studied quantum field theory, maybe it's
> > explained there...
>
>     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
From: Darwin123 on 29 Nov 2009 17:02
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.
From: BURT on 29 Nov 2009 17:12
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
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