photons and reflection

[RichD]

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

 

[Phil Allison]

"RichDope"

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....

** Think you need to read up on how reflection of light really works.

Cos what you describe above sounds more like how a light pumped gas laser
works.

http://en.wikipedia.org/wiki/QED_(book)



..... Phil

 

[Salmon Egg]

In article
<7d8aade9-635f-4ef9-9c04-ebafbc5b7a42@o13g2000vbl.googlegroups.com>,
RichD <r_delaney2001@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

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

--
An old man would be better off never having been born.

 

[Skywise]

RichD <r_delaney2001@yahoo.com> wrote in news:7d8aade9-635f-4ef9-9c04-
ebafbc5b7a42@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
--
http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism
Seismic FAQ: http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html
Quake "predictions": http://www.skywise711.com/quakes/EQDB/index.html
Sed quis custodiet ipsos Custodes?

 

[Androcles]

"RichD" <r_delaney2001@yahoo.com> wrote in message
news:7d8aade9-635f-4ef9-9c04-ebafbc5b7a42@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.

 

[nuny@bid.nes]

On Nov 27, 4:24 pm, RichD <r_delaney2...@yahoo.com> wrote:

According to the wave theory of light, angle of
incidence equals angle of reflection.  No problem,
in theory or fact.

Not just the relatively new wave theory. "Snell's" law and ray
theory go back to Alhazen.

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....

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.

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?

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.

Very roughly speaking.

I never studied quantum field theory, maybe it's
explained there...

Hope that helps.


Mark L. Fergerson

 

[John Devereux]

Skywise <into@oblivion.nothing.com> writes:

RichD <r_delaney2001@yahoo.com> wrote in news:7d8aade9-635f-4ef9-9c04-
ebafbc5b7a42@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

 

[George Herold]

On Nov 27, 7:24 pm, RichD <r_delaney2...@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.

 

[PD]

On Nov 27, 6:24 pm, RichD <r_delaney2...@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

 

[whit3rd]

On Nov 27, 4:24 pm, RichD <r_delaney2...@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.

 

[BURT]

On Nov 29, 11:29 am, whit3rd <whit...@gmail.com> wrote:

On Nov 27, 4:24 pm, RichD <r_delaney2...@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

 

[Darwin123]

On Nov 27, 7:24 pm, RichD <r_delaney2...@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.

 

[BURT]

On Nov 29, 1:22 pm, Darwin123 <drosen0...@yahoo.com> wrote:

On Nov 27, 7:24 pm, RichD <r_delaney2...@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

 

[Darwin123]

On Nov 27, 7:24 pm, RichD <r_delaney2...@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.

 

[BURT]

On Nov 29, 2:02 pm, Darwin123 <drosen0...@yahoo.com> wrote:

On Nov 27, 7:24 pm, RichD <r_delaney2...@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

 

[Rich Grise]

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

 

[BURT]

On Nov 30, 11:59 am, Rich Grise <richgr...@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

 

[Salmon Egg]

In article <pan.2009.11.30.19.59.55.233194@example.net>,
Rich Grise <richgrise@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.

 

[BURT]

On Nov 30, 6:25 pm, Salmon Egg <Salmon...@sbcglobal.net> wrote:

In article <pan.2009.11.30.19.59.55.233...@example.net>,
 Rich Grise <richgr...@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

 

[George Herold]

On Nov 29, 5:12 pm, BURT <macromi...@yahoo.com> wrote:

On Nov 29, 2:02 pm, Darwin123 <drosen0...@yahoo.com> wrote:





On Nov 27, 7:24 pm, RichD <r_delaney2...@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.