photons and reflection

Misapplying a theory to an observation then claiming you've disproved 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.

:->

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

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.

The mirror wouldn't work if that is true.

They're bound to their shell and quantum jump. Of course you are a moron.

Mitch Raemsch

BURT wrote in news:c581b031-362b-4b21-a413- snipped-for-privacy@j9g2000prh.googlegroups.com:

Prove it. I'm listening.

Brian

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BURT wrote in news:43c0344b-f497-4beb-bc9b- snipped-for-privacy@2g2000prl.googlegroups.com:

Thermodynamics.

Brian

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.

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.

You're dumb.

Mitch Raemsch

So, according to your hypothesis, if a mirror is absorbing and re-emitting, what cause the emission to have a particular angle?

Just curious...

Bob Masta DAQARTA v5.00 Data AcQuisition And Real-Time Analysis

Quantization is violated.

Mitch Raemsch

BURT wrote in news:e9749752-39a5-411f-b626- snipped-for-privacy@x25g2000prf.googlegroups.com:

You're a crank.

Brian

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Show me where I am wrong Brian and how you are right in thermodynamics.

Mitch Raemsch

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.

BURT wrote in news:3a7407c1-d14a-4d33-8a67- snipped-for-privacy@h40g2000prf.googlegroups.com:

I climbed Mount Everest. Prove I didn't.

Brian

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.

used You mean in near-field?

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D

If both voltage and currend are at nouht then presumably there are diael=E8ctric and diamagnetic bodies in medium to offset them such thas their potential yields somewise.

E, D, B, H -> S; D -> E, H -> B, D -> B, H -> E

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D

What a mutt--anafot=F2n. Two fot=F2ns meet and make a plasmonic beam.

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D

How would you know the ratio of writers?

None of them show fasis or h=FDsteresis.

-Aut

Berkeley spatial vice versa

Nescientist Androcles, the medium is the charge, not nothing.

empty as in not moot?

ield is wherever the conductors take it. In this motor is parallel with th= e shaft, curling around the iron, but also across the brushes. "Curl" is mi= sleading.<

Yes, "curling". If the loop were infinite, there'd be no reaction. So it's "E o( -B,/t,", where o( is proports.

t 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 pack= ets of energy from individual atoms need contain no aether with properties = of permittivity or permeability, these are properties of matter and not a r= equirement for the transmission of energy. All that is necessary and suffic= ient 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:

epts 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 a= xis. 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 pole= s.<

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; plasm=F2ns do them all. And there is no infinity for the univers isn't infinitely eld.

-Aut