photons and reflection
RichD
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
<7d8aade9-635f-4ef9...@o13g2000vbl.googlegroups.com>,
RichD <r_dela...@yahoo.com> wrote:
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.
alie...@gmail.com
> 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
Skywise
ebafbc...@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_dela...@yahoo.com> wrote in message
news:7d8aade9-635f-4ef9...@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.
John Devereux
> RichD <r_dela...@yahoo.com> wrote in news:7d8aade9-635f-4ef9-9c04-
> ebafbc...@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
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
> 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.
whit3rd
> 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
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
>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.
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?
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
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
> incidence equals angle of reflection. No problem,
> in theory or fact.
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?
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?
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?
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?
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...
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
No. There is no particle of light. It is easily demostratable as a
question that cannot be answered.
Mitch Raemsch
Rich Grise
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
>
> - 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
Rich Grise <rich...@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!
BURT
> 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
>
> - 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.
Louis Boyd
Neither wave or QM theory does a thorough job of explaining the observed
effects of electromagnetic energy interacting with matter. Wave theory
is generally more useful when dealing with propagation, refraction,
reflection and diffraction through and around material objects.
QM is generally more useful when electromagnetic energy interacts with
matter and energy is exchanged. They're both incomplete models of what
happens in nature. Use the one which works best to explain a given
phenomena. Or come up with more complete unified model if you can.
Lots of luck. It's not like others haven't tried with varying degrees
of success but the results are generally are too cumbersome to be useful.
Darwin123
> 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
> 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
I would like to hear you discuss photoluminescence and Raman
scattering!
Helpful person
>
> Neither wave or QM theory does a thorough job of explaining the observed
> effects of electromagnetic energy interacting with matter. Wave theory
> is generally more useful when dealing with propagation, refraction,
> reflection and diffraction through and around material objects.
> QM is generally more useful when electromagnetic energy interacts with
> matter and energy is exchanged. They're both incomplete models of what
> happens in nature. Use the one which works best to explain a given
> phenomena. Or come up with more complete unified model if you can.
> Lots of luck. It's not like others haven't tried with varying degrees
> of success but the results are generally are too cumbersome to be useful.
Thanks for the words of sanity!
alie...@gmail.com
Well, that explains lasers. No, wait, it doesn't.
> Mitch Raemsch - Still in the aether of time
Too bad you refuse to enter the real world.
Mark L. Fergerson
Rich Grise
> --
> An old man would be better off never having been born.
So, when do you plan to eat the golden bullet? 'Cause otherwise, you're
going to be one one day.
Good Luck!
Rich
BURT
>
> - Show quoted text -
But it doesn't explain a rainbow. Emission can be quantized but not
all of the time.
Mitch Raemsch
Salmon Egg
Rich Grise <rich...@example.net> wrote:
> So, when do you plan to eat the golden bullet? 'Cause otherwise, you're
> going to be one one day.
I am one already getting even oldfer.
Bill
p.ki...@ic.ac.uk
> What wave is the particle of light in? the electric opr
> magnetic wave?
Here's how the theory can be described (simplified, obviously):
(a) solve Maxwell's equations for a suitable system, and get a set
of normalizable basis functions allowing you to describe any field
configuration.
(b) these basis functions usually have both electric and magnetic
field contributions; they are usually called "mode functions", and
tend to oscillate in space and time (although not all will).
(c) quantize the field inside each mode; this gives you a countable
series of possible mode excitations.
(d) to describe some chosen field configuration, you combine a
suitable set of modes containing appropriate quantum excitations.
You may need to account for non-trivial correlations between the
modes, and between the quantum states in the same and different
modes.
There is no "particle of light". Instead there are countable
excitations of the wave-like field modes. These modes usually
combine both electric and magnetic contributions.
It's not a particle, it's a wave. But you _can_ count the
excitations.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (Photonics) (ph) +44-20-759-47734 (fax) 47714
Imperial College London, Dr.Paul...@physics.org
SW7 2AZ, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
George Herold
Louis, do you have any specific examples in mind? I thought the
theorists had a good handle on light. (I'm an experimentalist and am
certainly not going to come up with any theories of my own..... I've
got enough trouble understanding E&M, let alone QM.) When I measure
light it always comes as photons. With a Si Photodiode I get one
electron generated for each photon absorbed.
George H.
George Herold
> SW7 2AZ, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
Paul, I feel I'm in way over my head, but is there something wrong
with calling the excited quantized mode a photon?
George H.
AES
p.ki...@ic.ac.uk wrote:
> Here's how the theory can be described (simplified, obviously):
>
> (a) solve Maxwell's equations for a suitable system, and get a set
> of normalizable basis functions allowing you to describe any field
> configuration.
>
> (b) these basis functions usually have both electric and magnetic
> field contributions; they are usually called "mode functions", and
> tend to oscillate in space and time (although not all will).
>
> (c) quantize the field inside each mode; this gives you a countable
> series of possible mode excitations.
>
> (d) to describe some chosen field configuration, you combine a
> suitable set of modes containing appropriate quantum excitations.
> You may need to account for non-trivial correlations between the
> modes, and between the quantum states in the same and different
> modes.
>
>
> There is no "particle of light". Instead there are countable
> excitations of the wave-like field modes. These modes usually
> combine both electric and magnetic contributions.
>
> It's not a particle, it's a wave. But you _can_ count the
> excitations.
Agreed. That's absolutely the classic (not classical!) approach to
quantum analysis of e-m problems.
But any thoughts on how to extend this approach (or connect it) to the
whole class of "open systems" (all the systems like unstable
resonators/gain-guided waveguides, etc) where the actual operating or
oscillating or amplifying modes of the system (the "real modes", not
just some set of basis functions) are non-orthogonal, non-Hermitian,
non-self-adjoint, biorthogonal, so that you run into the Petermann
excess noise factor and "adjoint coupling" concepts and so on.
Obviously you can choose any more or less arbitrary set of Hermitian
basis functions to use in analyzing these systems; but since these
basis functions will _not_ be the actual "modes" that the system
actually operates in, your superposition will in general, and more or
less unavoidably, be a very inefficient way of describing the system.
I believe (and I think Han Woerdman does) that the nonhermitian
biorothogonal modes + Petermann excess noise factor approach predicts
the correct quantum results for these systems (or at least some of the
important quantum results?), and futhermore does so in an efficient and
simple fashion. But, I've never really understood how this approach
ties into, or can be connected to, the classic approach you describe.
BURT
> SW7 2AZ, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
There are only a very few quantizations in light energy quantities of
the atom. Certainly not enough for white light we see. This does not
correspond to the reality of the full spectrum produced by the white
light. A light bulb passed through a prism produces a full spectrum of
energy levels but does not have enough quantized states in its atom to
do so.
Mitch Raemsch
George Herold
>
> - Show quoted text -
Mitch, The light bulb can be thought of as a black body radiator. It
doesn't matter what kind of atoms the black body is made of. All that
is important is the temperature.
http://en.wikipedia.org/wiki/Blackbody_radiation
George H.
p.ki...@ic.ac.uk
> > It's not a particle, it's a wave. But you _can_ count the
> > excitations.
> Paul, I feel I'm in way over my head, but is there something wrong
> with calling the excited quantized mode a photon?
Calling a single (extra) excitation of a mode a "photon"
is pretty much exactly what you should do. Just don't
call it a particle as well.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (Photonics) (ph) +44-20-759-47734 (fax) 47714
Imperial College London, Dr.Paul...@physics.org
p.ki...@ic.ac.uk
>> > There is no "particle of light". Instead there are countable
> > excitations of the wave-like field modes. These modes usually
> > combine both electric and magnetic contributions.
> But any thoughts on how to extend this approach (or connect it) to the
> whole class of "open systems" (all the systems like unstable
> resonators/gain-guided waveguides, etc) where the actual operating or
> oscillating or amplifying modes of the system (the "real modes", not
> just some set of basis functions) are non-orthogonal, non-Hermitian,
> non-self-adjoint, biorthogonal, so that you run into the Petermann
> excess noise factor and "adjoint coupling" concepts and so on.
Oh, maybe (I never had cause to use this for anything though)
Brown S.A.; Dalton B.J.
http://arxiv.org/abs/quant-ph/0107039
... also in JMO 49, 1009 (2002)
http://dx.doi.org/10.1080/09500340110095625
BURT
>
> - Show quoted text -
George my point is that energy transitions cannot be quantized in the
case of a white light. You might have a light filliment composed of a
few different atoms but these could not produce the full spectrum of
all the light energies noticed when its light is passed through a
prism.
Evidently only sometimes is light energy quantized.
Mitch Raemsch
George Herold
Ahh, there are two types of quantization here. For an atom you have
quantized electron states. The photon emmited when the atom goes from
one state to the other has a particular 'quantized' frequency. But
this is just because of the uderlying quantized electron states.
There is then the quantization of the EM field that is called a
photon....(And I'll never call it a particle again.) When you measure
light you either see one photon or none....never some fraction of a
photon. (OK, most times you see lots of photons, but always an
interger number.)
George H.
(I was afraid you were going to ask, "From where comes the photon
emmited by a black body?" I don't have a good picture of that
process.)
BURT
>
> - Show quoted text -
The electron state is simply which of the 4 shells it is in. There are
only 4 fundamental sizes to the atom because of these round shells
that science calles energy levels of the electron.
White light from a surface composed of a few different atoms is
evidence that emmision is not always quantized.
Mitch Raemsch
Darwin123
> There is no "particle of light". Instead there are countable
> excitations of the wave-like field modes. These modes usually
> combine both electric and magnetic contributions.
However, in quantum mechanics the amplitude of this wave is
quantized. A wave with a quantized amplitude is fare different than a
wave that has an amplitude that can be contiuously varied.
>
> It's not a particle, it's a wave. But you _can_ count the
> excitations.
I don't think there is always a large difference between a wave
with quantized amplitude and a particle. If one has a "wave" at a high
quantization state, then I suppose it acts a bit like the classical
"wave". However, at small energy densities the energy has to become
somewhat localized.
I remember far back reading a mathematical analysis that showed
that a boson field with a quantized amplitude behave in all ways like
a "boson particle," except with regard to the ground state of the
excitation. The ground state of the boson excitation tends to have
"nonparticle" properties no matter how weak the field. However, at
quantum amplitudes that are not too low and not too high, a particle
description is valid. Hence, describing photons as a "particle" is
somewhat accurate.
I agree with your larger point. Photons are not classical
particles, and shouldn't be presented as such. Newton's corpuscular
theory is dead. However, his corpuscular theory is a -well- reanimated
corpse|;-)
Bill Taylor
The upper part represents the wave aspect;
the lower part represents the particle aspect.
-- Befuddled Bill
** They travel as waves but arrive as photons.
Androcles
"Bill Taylor" <w.ta...@math.canterbury.ac.nz> wrote in message
news:cdf336a9-30f5-4657...@z35g2000prh.googlegroups.com...
The upper part is the magnetic aspect;
the lower part is the electric aspect.
http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.htm
If a rotating magnet turns another magnet then there is
an energy transfer across empty space. One rotation
corresponds to one photon.
That's the nature of light.
BURT
Which wave is the particle in the the Electric or the Magnetic wave?
This question disproves the photon.
MItch Raemsch
Bob May
This defines an EQUIVALENT mass for an energy field. This doesn't meean
that a photon necessarily has mass but it does carry energy
--
Bob May
rmay at nethere.com
http: slash /nav.to slash bobmay
http: slash /bobmay dot astronomy.net
BURT
Refelection comes at every angle and every light energy in the
spectrum for white light. This means qunatization of energy coming out
of the atom isn't always applicable; for example the rainbow.
Mitch Raemsch
George Herold
Hmmm, Mitch if you are really interested in this stuff you'll have to
take a class (well there are QM classes in video on the web.) and work
through the standard problems. There are an infite number of energy
states of an atom. At the larger quantum numbers the states have
almost the same energy and we then just call them 'the continum' (or
some such term.) I have no idea where the 4 number came from but it's
wrong.
George H.
BURT
I don't read and I don't go to school but light still comes from every
angle in reflection.
Mitch Raemsch
BURT
A mirror is an example of a metal coating that can handle every
frequency of light. Quantization does not apply here either. A rainbow
and anything exibiting white light cannot be a phenomenon of
quantization of energy in the atom. A laser would be an exception that
needs to be taken into account. Evidently quantization has a limited
applicability.
Mitch Raemsch
George Herold
BURT wrote:
<big snip>
> > George H.- Hide quoted text -
> >
> > - Show quoted text -
>
> I don't read and I don't go to school but light still comes from every
> angle in reflection.
>
> Mitch Raemsch
"> I don't read and I don't go to school"
Oh well, as Newton (?) said, “I can see so far, because I’m standing
on the shoulders of giants. “
George H.
Louis Boyd
> A mirror is an example of a metal coating that can handle every
> frequency of light. Quantization does not apply here either. A rainbow
> and anything exibiting white light cannot be a phenomenon of
> quantization of energy in the atom. A laser would be an exception that
> needs to be taken into account. Evidently quantization has a limited
> applicability.
It's my understanding that quantization applies to the detection of EM
radiation when the detection involve changing electron states including
modifying chemical bonds. It's my understanding that it doesn't apply
when detection is accomplished simply by heating (increase in
molecular velocity) not involving ionization. So QM can apply to white
light and rainbows if you use your eye, film, or a CCD but not if you
use a bolometer or thermometer as the detector.
I could be dead wrong, but I consider photons to only "exist" when and
where light exchanges energy with matter at a sub-molecular level. At
least that view seems to be sufficient for engineering needs when
working with emitters and detectors.
Educate me. How does that view conflict with formal QM theory?
BURT
Evidently some atoms can radiate and absorb all visable frequencies
such as in the example of a mirror. I am not educated.
Mitch Raemsch
Bret Cannon
"Louis Boyd" <bo...@apt0.sao.arizona.edu> wrote in message
news:hfcnmb$6m4$1...@onion.ccit.arizona.edu...
"Photons" show up, that is light detection is appears and discrete events,
when light is detected, even with thermal detectors. To observe this
experimentally you of course need high enough signal to noise, but it is
done. For example, super conducting bolometers can not only detect soft
x-ray photons as discrete detection events, but also measure their energy
with a resolution of a few percent or better. I remember an interesting
talk on this work by a professor of physics or astronomy from the University
of Wisconsin at Madison.
Bret Cannon
BURT
> "Louis Boyd" <b...@apt0.sao.arizona.edu> wrote in message
>
> - Show quoted text -
If light is a particle which of its waves is this particle in? its
magnetic wave or electric wave?
No. Einstein questioned his photon in the end. He questioned what he
won the Nobel Prize for.
Reflection of a mirror is not quantized and happens over all visual
frequencies of light.
Quantization has been disproven. It does have application but it is
the lesser truth.
Mitch Raemsch
Skywise
525a12...@m7g2000prd.googlegroups.com:
> Quantization has been disproven. It does have application but it is
> the lesser truth.
So, please tell us. What is the greater truth?
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?
Bob Masta
<macro...@yahoo.com> wrote:
>If light is a particle which of its waves is this particle in? its
>magnetic wave or electric wave?
Yes. Don't think of this as "either-or", think of
the photon as the oscillation between magnetic and
electric fields. To use a mechanical analogy, you
might think of the photon as a rubber ball flying
through space. It is springy in the X and Y
dimensions, and oscillates between having its
energy stored in X-compression/Y-elongation,
versus Y-compression/X-elongation.
Now take away the ball.
Best regards,
Bob Masta
DAQARTA v5.00
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
Frequency Counter, FREE Signal Generator
Pitch Track, Pitch-to-MIDI
DaqMusic - FREE MUSIC, Forever!
(Some assembly required)
Science (and fun!) with your sound card!
Androcles
"Bob Masta" <N0S...@daqarta.com> wrote in message
news:4b1d0fb6...@news.sysmatrix.net...
> On Sun, 6 Dec 2009 20:31:18 -0800 (PST), BURT
> <macro...@yahoo.com> wrote:
>>If light is a particle which of its waves is this particle in? its
>>magnetic wave or electric wave?
>
> Yes. Don't think of this as "either-or", think of
> the photon as the oscillation between magnetic and
> electric fields. To use a mechanical analogy, you
> might think of the photon as a rubber ball flying
> through space. It is springy in the X and Y
> dimensions, and oscillates between having its
> energy stored in X-compression/Y-elongation,
> versus Y-compression/X-elongation.
>
> Now take away the ball.
>
> Best regards,
>
>
> Bob Masta
Quite right, Bob. I sometimes use a "leapfrog" analogy, the electric
field creates the magnetic field as the electric collapses and then the
magnetic creates the electric field in turn.
http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.htm
BURT
> BURT <macromi...@yahoo.com> wrote in news:b72906b1-e886-418b-8f3e-
> 525a12b37...@m7g2000prd.googlegroups.com:
>
> > Quantization has been disproven. It does have application but it is
> > the lesser truth.
>
> So, please tell us. What is the greater truth?
>
> Brian
> Seismic FAQ:http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html
> Quake "predictions":http://www.skywise711.com/quakes/EQDB/index.html
> Sed quis custodiet ipsos Custodes?
Quantization is less imorportant.
Mitch Raemsch
Skywise
2fb244...@z35g2000prh.googlegroups.com:
> On Dec 6, 10:43�pm, Skywise <i...@oblivion.nothing.com> wrote:
>> BURT <macromi...@yahoo.com> wrote in news:b72906b1-e886-418b-8f3e-
>> 525a12b37...@m7g2000prd.googlegroups.com:
>>
>> > Quantization has been disproven. It does have application but it is
>> > the lesser truth.
>>
>> So, please tell us. What is the greater truth?
>>
> Quantization is less imorportant.
So which is it? "less important" or "has been disproven"?
One implies it exists and the other that it doesn't.
Brian
--
http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism
BURT
> BURT <macromi...@yahoo.com> wrote in news:abdfaf69-fc01-40fe-a43e-
> 2fb244fc2...@z35g2000prh.googlegroups.com:
>
> > On Dec 6, 10:43 pm, Skywise <i...@oblivion.nothing.com> wrote:
> >> BURT <macromi...@yahoo.com> wrote in news:b72906b1-e886-418b-8f3e-
> >> 525a12b37...@m7g2000prd.googlegroups.com:
>
> >> > Quantization has been disproven. It does have application but it is
> >> > the lesser truth.
>
> >> So, please tell us. What is the greater truth?
>
> > Quantization is less imorportant.
>
> So which is it? "less important" or "has been disproven"?
>
> One implies it exists and the other that it doesn't.
>
> Brian
> Seismic FAQ:http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html
> Quake "predictions":http://www.skywise711.com/quakes/EQDB/index.html
> Sed quis custodiet ipsos Custodes?
The truth is meant to be known. Quantization is the lesser concept
than full ranges of energy.
Mitch Raemsch
p.ki...@ic.ac.uk
> Quite right, Bob. I sometimes use a "leapfrog" analogy, the electric
> field creates the magnetic field as the electric collapses and then the
> magnetic creates the electric field in turn.
Why not consider the Yee grid for discretizing and numerically solving
Maxwell's equations? That might be leapfroggy enough for you, without
requiring any dubious analogies.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (Photonics) (ph) +44-20-759-47734 (fax) 47714
Imperial College London, Dr.Paul...@physics.org
p.ki...@ic.ac.uk
> The truth is meant to be known. Quantization is the lesser
> concept than full ranges of energy.
The process of quantization doesn't require discrete, countable
sets of EM mode functions (although that may well make the
mathematics easier). Quantization doesn't necessarily restrict
the allowed energies.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (Photonics) (ph) +44-20-759-47734 (fax) 47714
Imperial College London, Dr.Paul...@physics.org
p.ki...@ic.ac.uk
> Refelection comes at every angle and every light energy in the
> spectrum for white light. This means qunatization of energy coming out
> of the atom isn't always applicable; for example the rainbow.
You miss the point: I can quantize the field in any set of basis
modes I happen to prefer. However, some basis sets will provide
simpler descriptions of the behavior than others. And for some
situations, no description will be simple, and I'll have to settle
for least-complicated instad.
The process of "Quantizing the field" isn't unique. It's a choice.
Generally, though, there is only a small set of useful bases for
which quantization gives a useful description.
eric gisse
> BURT <macro...@yahoo.com> wrote:
>> The truth is meant to be known. Quantization is the lesser
>> concept than full ranges of energy.
>
> The process of quantization doesn't require discrete, countable
> sets of EM mode functions (although that may well make the
> mathematics easier). Quantization doesn't necessarily restrict
> the allowed energies.
>
You are wasting your time explaining this to him. He is nothing but a noise
generator.
Androcles
<p.ki...@ic.ac.uk> wrote in message
news:4aq1v6-...@ph-kinsle.qols.ph.ic.ac.uk...
> Androcles <Headm...@hogwarts.physics_q> wrote:
>> Quite right, Bob. I sometimes use a "leapfrog" analogy, the electric
>> field creates the magnetic field as the electric collapses and then the
>> magnetic creates the electric field in turn.
>
> Why not consider the Yee grid for discretizing and numerically solving
> Maxwell's equations? That might be leapfroggy enough for you, without
> requiring any dubious analogies.
>
not play mathematical games. Maxwell plagiarised the work of Faraday,
Gauss and Ampere. Maxwell's aether was a dubious analogy dispelled
by Michelson, that's why.
George Herold
>
> - Show quoted text -
Hi Androcles. I use to like this analogy too. Until I learned a few
years ago that in E-M radiation the E and B are in phase! At first I
thought there was a mistake... but then discovered that the mistake
was mine. (Your link shows correctly the in phase behavior so I
realize I'm not telling you anything you don't know.)
Anyway the analogy can lead to false conclusions. (At least for me.)
So now I see that the E field at some time was 'created' by some B
field at a previous time.... Which starts to 'weird' me out if think
too hard. All of a sudden I picture 'photons' travelling in both
directions.
George H.
Androcles
"George Herold" <gghe...@gmail.com> wrote in message
news:0497f18c-647c-4792...@u7g2000yqm.googlegroups.com...
===========================================
Then you should unlearn it immediately. If E and B were in phase
both would be zero simultaneously and that violates the first law of
thermodynamics, you'd create energy from nothing.
http://en.wikipedia.org/wiki/First_law_of_thermodynamics
But anyway, Maxwell's equations never did claim E and B were in phase,
what you've "learnt" is a rumour spread by the incompetent.
===========================================
At first I
thought there was a mistake... but then discovered that the mistake
was mine. (Your link shows correctly the in phase behavior so I
realize I'm not telling you anything you don't know.)
Anyway the analogy can lead to false conclusions. (At least for me.)
So now I see that the E field at some time was 'created' by some B
field at a previous time....
============================================
Any spark will start the process. A flame is a chemical reaction
whereby the electrons of the atoms are rearranged to build a different
molecule. 2H2 + O2 -> 2H2O.
============================================
Which starts to 'weird' me out if think
too hard. All of a sudden I picture 'photons' travelling in both
directions.
George H.
===========================================
Androcles' third law: For every photon there is an equal and
opposite rephoton.
http://www.androcles01.pwp.blueyonder.co.uk/rephoton.gif
(It's Newton's third law applied to E-M waves and allows
for light to travel in beams -- quite simple, really.)
Of course you'll never see a rephoton without a mirror, it is
travelling away from you. Rephotons are the major cause of
poorly understood spooky entanglement.
--
'By denying scientific principles, one may maintain any paradox.' - Galileo
Galilei
'There is nothing so easy but that it becomes difficult when you do it with
reluctance.'- Marcus Tullius Cicero
New ideas are old ideas resurrected. - Androcles.
NoEinstein
>
Dear RichD: You are most correct that photons don't reflect (even
from the shiniest mirror). All photons get absorbed by the surface
and the electron orbits re emit light of the correct wave length.
Matter in the 'reflecting' surface doesn't need to KNOW the angle of
the mirror face. All of the atoms near the surface act to absorb the
"off angle" photons being re emitted such that only those photons
which have your stated angle can escape the surface. — NoEinstein —
>
> 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
BURT
> BURT <macromi...@yahoo.com> wrote:
> > The truth is meant to be known. Quantization is the lesser
> > concept than full ranges of energy.
>
> The process of quantization doesn't require discrete, countable
> sets of EM mode functions (although that may well make the
> mathematics easier). Quantization doesn't necessarily restrict
> the allowed energies.
Quantization is defined as making energy transitions discrete for the
electron.
It does not accomadate a mirror.
Show me where I am wrong.
Mitch Raemsch
>
> --
> ---------------------------------+---------------------------------
> Dr. Paul Kinsler
> Blackett Laboratory (Photonics) (ph) +44-20-759-47734 (fax) 47714
> Imperial College London, Dr.Paul.Kins...@physics.org
George Herold
> "George Herold" <ggher...@gmail.com> wrote in message
>
> news:0497f18c-647c-4792...@u7g2000yqm.googlegroups.com...
> On Dec 7, 10:53 am, "Androcles" <Headmas...@Hogwarts.physics_q> wrote:
>
>
>
>
>
> > "Bob Masta" <N0S...@daqarta.com> wrote in message
>
> >news:4b1d0fb6...@news.sysmatrix.net...
>
> > > On Sun, 6 Dec 2009 20:31:18 -0800 (PST), BURT
> > > <macromi...@yahoo.com> wrote:
> > >>If light is a particle which of its waves is this particle in? its
> > >>magnetic wave or electric wave?
>
> > > Yes. Don't think of this as "either-or", think of
> > > the photon as the oscillation between magnetic and
> > > electric fields. To use a mechanical analogy, you
> > > might think of the photon as a rubber ball flying
> > > through space. It is springy in the X and Y
> > > dimensions, and oscillates between having its
> > > energy stored in X-compression/Y-elongation,
> > > versus Y-compression/X-elongation.
>
> > > Now take away the ball.
>
> > > Best regards,
>
> > > Bob Masta
>
> > Quite right, Bob. I sometimes use a "leapfrog" analogy, the electric
> > field creates the magnetic field as the electric collapses and then the
> > magnetic creates the electric field in turn.
>
> - Show quoted text -
Hmm, Sorry on my second look your picture of a photon has it wrong
you've got the E and B fields 90 degrees out of phase. This is
exactly what I would have drawn a few years ago.
But check out this,
http://en.wikipedia.org/wiki/Electromagnetic_radiation
There is a picture if you scroll down a bit.
George H.
BURT
>
> There is a picture if you scroll down a bit.
>
>
> - Show quoted text -
Quantization doesn't work for rainbow physics.
Mitch Raemsch
Androcles
=====================================
It's right. Just ask any electrical engineer.
=====================================
This is
exactly what I would have drawn a few years ago.
=====================================
You'd have been right years ago.
=====================================
But check out this,
http://en.wikipedia.org/wiki/Electromagnetic_radiation
There is a picture if you scroll down a bit.
George H.
=================================================
Wackypedia is written by both incompetent kooks and the wise.
Kooks outnumber the wise by at least 100:1, perhaps a 1000:1.
Wackypedia has it wrong. See the discussion page, there are a set of
tabs labelled "article", "discussion", "edit this page" and "history"
at the top.
YOU can edit the page, I refuse to have anything to do with it.
Faraday wrote E = -dB/dt.
He did not write E = B, he did not write dE/dt = -dB/dt and
he did experiment. A CHANGING magnetic field produces
an electric field. Ask any generator designer.
The kook diagram you've indicated shows E = B.
Use this instead:
http://en.wikipedia.org/wiki/Trigonometric_functions
And do not write
- Hide quoted text -- Show quoted text -- Hide quoted text -- Show quoted
text -- Hide quoted text -- Show quoted text -- Hide quoted text -- Show
quoted text -, it irritates me.
Delete it before you post to usenet.
BURT
>
> - Show quoted text -
If the photon is in light which wave is it in?
Mitch Raemsch
Skywise
3363ec...@b25g2000prb.googlegroups.com:
> Show me where I am wrong.
I have climbed Mount Everest. Prove me wrong.
Brian
--
http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism
Skywise
a67ed3...@2g2000prl.googlegroups.com:
> The truth is meant to be known. Quantization is the lesser concept
> than full ranges of energy.
So then, what is the truth? If quantization is 'less correct', then
what is 'more correct'. If you're going to tell us we're wrong, then
tell us what's right. We're listening.
Brian
--
http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism
Bob Masta
<macro...@yahoo.com> wrote:
>If the photon is in light which wave is it in?
Why do you think it needs to be "in" one wave or
the other?
George Herold
> "George Herold" <ggher...@gmail.com> wrote in message
>
> news:bc2be262-2a4f-4ba9...@b15g2000yqd.googlegroups.com...
> On Dec 8, 11:16 am, "Androcles" <Headmas...@Hogwarts.physics_q> wrote:
>
<snip>
Hi Androcles, Are you so sure you are correct? Or is there some
chance you could learn something new? I don't mind trying to work
through the mathematics with you.... But only if you are interested.
I know about Faraday's law. But we are talking about someting
different here. It is the travleing wave solution of Maxwells
equations. I think that in the near field of the source you will find
that the E and B fields are out of phase. But the far-field traveling
wave is different. I'm not much of a theorist. But looking over the
solutions (At the moment I'm looking at Volume 3 (Waves) of the
Berkley series on Physics) One can see that the spacial derivative of
B is equal to 1/c times the time derivative of E. (and visa versa.)
From which (with a little math) one can see that the E and B are in
phase. (Oh this is the free space solution.)
George H.
Androcles
<snip>
=================================
Yes, quite sure.
=================================
Or is there some
chance you could learn something new?
=================================
I strongly doubt you have any new evidence or data,
and I'm not really interested in old crackpot theories.
But present it if you do.
=================================
I don't mind trying to work
through the mathematics with you.... But only if you are interested.
==================================
I've already presented
http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.htm
which you've snipped.
If you think you can fault it you are welcome to try. Just
remember that mathematics is about proof based on axioms.
==================================
I know about Faraday's law.
==================================
That's nice for you.
==================================
But we are talking about someting
different here.
=================================
No we are not. We are talking about the transfer of energy
through nothing, as when you feel the heat and see a big
bright ball in the sky.
=================================
It is the travleing wave solution of Maxwells
equations.
=================================
What travelling wave and what is waving? I've
seen no evidence of a travelling wave.
I think
=================================
You can stop right there. I'm not interested in what you
think, show we what you can prove.
=================================
that in the near field of the source you will find
that the E and B fields are out of phase.
But the far-field traveling
wave is different. I'm not much of a theorist.
=================================
I'm not interested in your theories. You said above
you'd work through the mathematics. I'll allow that,
but I'm not going to listen to your theories.
=================================
But looking over the
solutions (At the moment I'm looking at Volume 3 (Waves) of the
Berkley series on Physics) One can see that the spacial derivative of
B is equal to 1/c times the time derivative of E. (and visa versa.)
From which (with a little math) one can see that the E and B are in
phase. (Oh this is the free space solution.)
=================================
Phase shift is found in the time derivative. It's E = -dB/dt,
and E <> -dB/dx.
Or is there some chance you could learn something old?
BURT
> BURT <macromi...@yahoo.com> wrote in news:8b79ee65-5af0-44a2-bf00-
> a67ed367a...@2g2000prl.googlegroups.com:
>
> > The truth is meant to be known. Quantization is the lesser concept
> > than full ranges of energy.
>
> So then, what is the truth? If quantization is 'less correct', then
> what is 'more correct'. If you're going to tell us we're wrong, then
> tell us what's right. We're listening.
>
> Brian
> Seismic FAQ:http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html
> Quake "predictions":http://www.skywise711.com/quakes/EQDB/index.html
> Sed quis custodiet ipsos Custodes?
Quantization only applies to stimulated emmision.
Opaque objects have to absorb all frequencies.
Quantum Mechanics is wrong.
Mitch Raemsch
PD
> On Dec 8, 8:26 pm, Skywise <i...@oblivion.nothing.com> wrote:
>
> > BURT <macromi...@yahoo.com> wrote in news:8b79ee65-5af0-44a2-bf00-
> > a67ed367a...@2g2000prl.googlegroups.com:
>
> > > The truth is meant to be known. Quantization is the lesser concept
> > > than full ranges of energy.
>
> > So then, what is the truth? If quantization is 'less correct', then
> > what is 'more correct'. If you're going to tell us we're wrong, then
> > tell us what's right. We're listening.
>
> > Brian
> > Seismic FAQ:http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html
> > Quake "predictions":http://www.skywise711.com/quakes/EQDB/index.html
> > Sed quis custodiet ipsos Custodes?
>
> Quantization only applies to stimulated emmision.
Oh dear.
BURT
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Perhaps you have something to say?
Mitch Raemsch
PD
Oh, I dunno. Some statements you make are just so far off base,
there's too much to correct.
You might have well have said that mathematics only applies to adding
up the prices of groceries.
BURT
is absorption and emmision taking place accross all visual
frequencies. Quantum mechanics is wrong about quantization of energy
levels for a mirror's atoms. It doesn't work. The broader truth is not
quantization and quantum mechanics needs to be corrected.
Mitch Raemsch
Louis Boyd
the theory is just silly. Quantum mechanics is useful. QM is useful
when describing the interaction of light and matter at a sub molecular
level If QM was expanded to explain all wave phenomena it would be too
cumbersome to use. What's so difficult about just applying wave and QM
when they're needed? Most people never need either theory. You can sit
on a couch and watch football games on TV without any knowledge of
either theory. Theories are simply tools. If a hammer works and a
screwdriver doesn't then use the hammer. You could make a tool with a
hammer on one end an a screwdriver on the other but it wouldn't be as
useful as two separate tools.
Jamie
BURT
Einstein said that Quantum Mechanics was wrong because God did not
play dice with the universe.
Evidently it is also wrong for a mirror.
Mitch Raemsch
whit3rd
> A mirror is an example that dispoves quantum mechanics. The reflection
> is absorption and emmision taking place accross all visual
> frequencies. Quantum mechanics is wrong about quantization of energy
> levels for a mirror's atoms. It doesn't work.
The mirror reflects only if the light incoming is under the plasma
frequency of the free electrons in the metallic reflector layer (there
are other reflection mechanisms for nonmetals). The reflection is
equal to the incoming light, so there is NO 'transition' between
energy states required in any particle.
No absorption. No emission. The metallic bonding
of the atoms in the mirror's silver layer acts to free the electrons
from those atomic energy-level rules, in this case. You can't
make a normal mirror from un-attached atoms (gases) because
the electrons are bound and not free.
BURT
> On Dec 10, 11:40 am, BURT <macromi...@yahoo.com> wrote:
>
> > A mirror is an example that dispoves quantum mechanics. The reflection
> > is absorption and emmision taking place accross all visual
> > frequencies. Quantum mechanics is wrong about quantization of energy
> > levels for a mirror's atoms. It doesn't work.
>
> The mirror reflects only if the light incoming is under the plasma
> frequency of the free electrons in the metallic reflector layer (there
> are other reflection mechanisms for nonmetals). The reflection is
> equal to the incoming light, so there is NO 'transition' between
> energy states required in any particle.
>
> No absorption. No emission.
The mirror wouldn't work if that is true.
> The metallic bonding
> of the atoms in the mirror's silver layer acts to free the electrons
> from those atomic energy-level rules, in this case. You can't
> make a normal mirror from un-attached atoms (gases) because
> the electrons are bound and not free.
They're bound to their shell and quantum jump. Of course you are a
moron.
Mitch Raemsch
Skywise
df5644...@j9g2000prh.googlegroups.com:
> Quantum Mechanics is wrong.
Prove it. I'm listening.
Brian
--
http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism
Skywise
7bd8f0...@2g2000prl.googlegroups.com:
Thermodynamics.
whit3rd
> On Dec 10, 6:11 pm, whit3rd <whit...@gmail.com> wrote:
>
> > On Dec 10, 11:40 am, BURT <macromi...@yahoo.com> wrote:
> > ... there is NO 'transition' between
> > energy states required in any particle.
>
> > No absorption. No emission.
>
> The mirror wouldn't work if that is true.
The interaction of light and matter is not exclusively by atomic
absorption
and emission. Neither atomic absorption nor emission is necessary to
model a mirror's action.
>
> > The metallic bonding
> > of the atoms in the mirror's silver layer acts to free the electrons
> They're bound to their shell and quantum jump.
Shells are for atoms. In solid or liquid metals, it's 'energy bands'.
Because the conduction band of conductive metals is not full of
electrons, an
individual electron can accelerate while remaining inside the band.
That's why free electrons are important. Conduction of electricity
and
reflection of light both require those free electrons in a wide
energy band.
BURT
You're dumb.
Mitch Raemsch
Bob Masta
So, according to your hypothesis, if a mirror is
absorbing and re-emitting, what cause the emission
to have a particular angle?
Just curious...
BURT
> On Thu, 10 Dec 2009 11:40:26 -0800 (PST), BURT
>
Quantization is violated.
Mitch Raemsch
Skywise
b2dc1d...@x25g2000prf.googlegroups.com:
> You're dumb.
You're a crank.
BURT
> BURT <macromi...@yahoo.com> wrote in news:e9749752-39a5-411f-b626-
> b2dc1dbee...@x25g2000prf.googlegroups.com:
>
> > You're dumb.
>
> You're a crank.
>
> Brian
> Seismic FAQ:http://www.skywise711.com/SeismicFAQ/SeismicFAQ.html
> Quake "predictions":http://www.skywise711.com/quakes/EQDB/index.html
> Sed quis custodiet ipsos Custodes?
Show me where I am wrong Brian and how you are right in
thermodynamics.
Mitch Raemsch
Anti Vigilante
Um, uh huh. No! Objects must reflect or allow through what they cannot
receive.
They cannot receive that which has no room to be contained.
They cannot receive half or 1 and half of a full cycle from its point of
view because they would remain perturbed and the perturbation would snap
back and spit out the radiation.
> Quantum Mechanics is wrong.
>
> Mitch Raemsch
--
Fuck the Enlightenment! Viva la Renaissance!
Skywise
7c526d...@h40g2000prf.googlegroups.com:
I climbed Mount Everest. Prove I didn't.
Brian
--
http://www.skywise711.com - Lasers, Seismology, Astronomy, Skepticism
Autymn D. C.
> On Dec 8, 3:39 am, p.kins...@ic.ac.uk wrote:
>
> > BURT <macromi...@yahoo.com> wrote:
> > > The truth is meant to be known. Quantization is the lesser
> > > concept than full ranges of energy.
>
> > sets of EM mode functions (although that may well make the
> > mathematics easier). Quantization doesn't necessarily restrict
> > the allowed energies.
>
> Quantization is defined as making energy transitions discrete for the
> electron.
> It does not accomadate a mirror.
>
> Show me where I am wrong.
A networkling or molecular band material has finite bodies and finite
lifetime; therefore it has finite transitions. Continvum radiation is
a figure of speech for a spectrum's thorouhness and there's no
threshhold where a spectrum becomes sheer--"sheer" also a relative
term.
Autymn D. C.
> "George Herold" <ggher...@gmail.com> wrote in message
> news:bc2be262-2a4f-4ba9...@b15g2000yqd.googlegroups.com...
> On Dec 8, 11:16 am, "Androcles" <Headmas...@Hogwarts.physics_q> wrote:
> > "George Herold" <ggher...@gmail.com> wrote in message
> > Hi Androcles. I use to like this analogy too. Until I learned a few
> > years ago that in E-M radiation the E and B are in phase!
used
You mean in near-field?
> > ===========================================
> > Then you should unlearn it immediately. If E and B were in phase
> > both would be zero simultaneously and that violates the first law of
> > thermodynamics, you'd create energy from nothing.
> >http://en.wikipedia.org/wiki/First_law_of_thermodynamics
> > But anyway, Maxwell's equations never did claim E and B were in phase,
> > what you've "learnt" is a rumour spread by the incompetent.
> > ===========================================
If both voltage and currend are at nouht then presumably there are
diaelèctric and diamagnetic bodies in medium to offset them such thas
their potential yields somewise.
> > At first I
> > thought there was a mistake... but then discovered that the mistake
> > was mine. (Your link shows correctly the in phase behavior so I
> > realize I'm not telling you anything you don't know.)
>
> > Anyway the analogy can lead to false conclusions. (At least for me.)
>
> > So now I see that the E field at some time was 'created' by some B
> > field at a previous time....
E, D, B, H -> S; D -> E, H -> B, D -> B, H -> E
> > ============================================
> > Any spark will start the process. A flame is a chemical reaction
> > whereby the electrons of the atoms are rearranged to build a different
> > molecule. 2H2 + O2 -> 2H2O.
> > ============================================
>
> > Which starts to 'weird' me out if think
> > too hard. All of a sudden I picture 'photons' travelling in both
> > directions.
>
> > George H.
>
> > ===========================================
> > Androcles' third law: For every photon there is an equal and
> > opposite rephoton.
> >http://www.androcles01.pwp.blueyonder.co.uk/rephoton.gif
> > (It's Newton's third law applied to E-M waves and allows
> > for light to travel in beams -- quite simple, really.)
What a mutt--anafotòn. Two fotòns meet and make a plasmonic beam.
> http://en.wikipedia.org/wiki/Electromagnetic_radiation
>
> There is a picture if you scroll down a bit.
>
> George H.
> =================================================
> Wackypedia is written by both incompetent kooks and the wise.
> Kooks outnumber the wise by at least 100:1, perhaps a 1000:1.
> Wackypedia has it wrong. See the discussion page, there are a set of
> tabs labelled "article", "discussion", "edit this page" and "history"
> at the top.
> YOU can edit the page, I refuse to have anything to do with it.
How would you know the ratio of writers?
> Faraday wrote E = -dB/dt.
> He did not write E = B, he did not write dE/dt = -dB/dt and
> he did experiment. A CHANGING magnetic field produces
> an electric field. Ask any generator designer.
None of them show fasis or hýsteresis.
-Aut
Autymn D. C.
spatial
vice versa
Nescientist Androcles, the medium is the charge, not nothing.
Autymn D. C.
> "Bill Taylor" <w.tay...@math.canterbury.ac.nz> wrote in message
>
> news:cdf336a9-30f5-4657...@z35g2000prh.googlegroups.com...
>
> > The nature of light is "?" .
>
> > The upper part represents the wave aspect;
> > the lower part represents the particle aspect.
>
> > -- Befuddled Bill
>
> > ** They travel as waves but arrive as photons.
>
> The upper part is the magnetic aspect;
> the lower part is the electric aspect.
>
> http://www.androcles01.pwp.blueyonder.co.uk/AC/AC.htm
>
> If a rotating magnet turns another magnet then there is
> an energy transfer across empty space. One rotation
> corresponds to one photon.
> That's the nature of light.
empty as in not moot? http://google.com/groups?q=%22Comparisons+for+the+illiterate%22
(update: find|leave, cleave|clive, meld|sunder, meet|split)
>Some purists want that to read curl E = -dB/dt but really the electric field is wherever the conductors take it. In this motor is parallel with the shaft, curling around the iron, but also across the brushes. "Curl" is misleading.<
Yes, "curling". If the loop were infinite, there'd be no reaction.
So it's "E o( -B,/t,", where o( is proports.
>When we do that we say we have a dielectric and a diamagnetic material that can affect the operation, but it is not a requirement for the process to occur. Thus the vast reaches of space across which light reaches us in packets of energy from individual atoms need contain no aether with properties of permittivity or permeability, these are properties of matter and not a requirement for the transmission of energy. All that is necessary and sufficient is that magnetic and electric fields must exist in the vacuum of space.<
The vacvum is still a material medium, the far-field of the radiant
body--namely, its charges. There is no transmission without matter:
http://google.com/groups?q=Autymn+-autumn+sun+bird.
>The "wave" nature of the photon is simply a misinterpretation of such concepts as "wavelength", for the wave shown above is a wave not in space, but in time. "Now" is shown by the black vertical line and as time passes the trace shows the voltage and gaussage* as it once was, not how it is now. It doesn't actually exist "now", but it did "then". There is no wave"length", only wave duration. The horizontal axis is the time axis, not a distance axis. The "poles" of the photon are it's centre and the surface of a sphere at infinity, for there is no electric or magnetic field except between poles.<
its
Length is in time. A wave (especially in condensed matter) has bobble
(room) and ripple (time) componends: wavearm (stride) and wavestint
(tide) in near-field/pole and wavespan (stride) and wavelength (tide)
in far-field/group; plasmòns do them all. And there is no infinity
for the univers isn't infinitely eld.
-Aut