Mike Raiford wrote:

> They are, sort of. In the real world, highlights would be caused by very 
> fine irregularities on the surface. which will invariably reflect some 
> of the light from the light source to your eye. What you see, then is a 
> diffused reflection of the object's surroundings. Some of which may 
> actually be a light source.
> 
> Highlights as POV-Ray defines them is a sort of short cut.

Right. This is basically what I thought. ;-)

Doesn't that mean that, in the real world, the hilights should be the 
shape of whatever luminous object is creating them? (The real world 
doesn't have point-lights.)

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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Orchid XP v8 wrote:
>> Highlights as POV-Ray defines them is a sort of short cut.
> 
> Right. This is basically what I thought. ;-)
> 
> Doesn't that mean that, in the real world, the hilights should be the 
> shape of whatever luminous object is creating them? (The real world 
> doesn't have point-lights.)

Aye. With additional deformation due to the shape of the surface, of course.

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Mike Raiford wrote:
> They are, sort of. In the real world, highlights would be caused by very 
> fine irregularities on the surface. 

You don't even need fine irregularities. Quantum randomness would be enough.

Conductive metal reflects shiney because there's a "layer" of free electrons 
floating between atoms, and they interact with the light in a much 
"smoother" way than electrons interacting with electrons bound to atoms. So 
being conductive is caused by the same thing as being reflectively shiny.

Of course, if the surface isn't smooth to start with, you're unlikely to see 
a mirror-like reflection to start with. But you're going to have a hard time 
polishing untreated wood to a shine no matter how smooth you make it, and if 
you start getting something like fluorescence, you start seeing that 
reflections aren't really "bouncing" light at all, but light that's absorbed 
and re-emitted..

-- 
   Darren New, San Diego CA, USA (PST)
   The NFL should go international. I'd pay to
   see the Detroit Lions vs the Roman Catholics.

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>> Highlights as POV-Ray defines them is a sort of short cut.
>
> Right. This is basically what I thought. ;-)
>
> Doesn't that mean that, in the real world, the hilights should be the 
> shape of whatever luminous object is creating them? (The real world 
> doesn't have point-lights.)

Yes, if your surface was perfectly smooth.  But IRL most surfaces are not, 
they have a roughness even if it is at a very small scale, thus the "normal" 
you use in the reflection equation is perturbed slightly which can make the 
highlight look larger and more blurred than it would due to a straight 
reflection.

http://en.wikipedia.org/wiki/Specular_highlight

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Darren New <dne### [at] sanrrcom> wrote:
> Of course, if the surface isn't smooth to start with, you're unlikely to see
> a mirror-like reflection to start with. But you're going to have a hard time
> polishing untreated wood to a shine no matter how smooth you make it

That's because wood in itself has such an extreme inhomogenity at microscopic
scale.

You can polish stone or plastic though to get mirror-like reflections.
They'll not be as prominent as in metal because they have to compete with
diffuse reflection, but the mirror-like component is still in there.


> and if
> you start getting something like fluorescence, you start seeing that
> reflections aren't really "bouncing" light at all, but light that's absorbed
> and re-emitted..

That's pure nonsense. If you deal with fluorescence you'll see that *all*
re-emitted light is *undirected* (at least with relation to the direction of
the absorbed light).

*Specular* reflections (i.e. the thing we'd normally call reflections - not to
be confused with PoV-ray's shortcut of "specular highlights") are *really*
"bouncing" light; it's the result of the electromagnetic wave (the "probability
wave" of the photon if you want to go Quantum) being unable to fully enter the
medium (speaking of permeability and permittivity), so the light wave (or part
of it) simply reflects, in order to comply with Maxwell's equations.

It can't work with absorption and re-emission - because the "incoming angle =
outgoing angle" law of specular reflection is a result of a light wave's
interference with itself. But as soon as a photon is absorbed by an electron,
the photon's probability wave collapses, so the re-emitted light's probability
wave has no way of interferencing with it, even if the direction of emission
would be in any way related to the absorbed photon.

It also can't work with absorption and re-emission because both are limited to
certain wavelengths for many materials, unless you go high temperature. Yet
specular reflections off most polished materials (except metals) are pure
"white" - even for materials that do show a strong color.

Only *Diffuse* reflections (i.e. the thing we'd normally not call reflections at
all) *may* sometimes be due to absorbed and re-emitted light.

Note however that quite a lot of instances of diffuse reflection are still
"bouncing" light. If you take a heap of fine-grained sugar for example, it the
diffuse white reflection is actually due to a very random sequence of
reflections and refractions at the single sugar crystals. A single "white"
sugar crystal is quite transparent though, so no absorption happening there.
And "brown" sugar gets its color because some of the light traversing the
crystals is absorbed by natural impurities - but there's no re-emission
involved (at least not to any significant degree in the visible spectrum).

You also typically get re-emissions at different (usually longer) wavelengths
than the absorbed photons (if only because of the principle of entropy). UV in,
visible out is a typical thing; or visible in, infrared out, though that's
usually due to thermal emissions. Green in, amber out is a possiblity as well.

In fact, what we'd normally call "diffuse" reflection is typically a combination
of some light "bouncing" at the surface, some light being absorbed by the
medium, some light "bouncing" below the surface ("subsurface scattering"), and
some (though usually not much) light (re-) emitted.

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clipka wrote:
> You can polish stone or plastic though to get mirror-like reflections.

You'll get recognisable reflections. You won't get a reflection like a 
mirror gives. That's more what I meant. You'll get "polished" rather than 
"shiny", if that makes sense.

> That's pure nonsense. If you deal with fluorescence you'll see that *all*
> re-emitted light is *undirected* (at least with relation to the direction of
> the absorbed light).

Of course this is true with flourescence, because you're emitting a 
different frequency than you absorbed. It's not true in general. See, for 
example, a laser, where all re-emitted light is specifically in the same 
direction and phase as the absorbed photon that triggers the re-emission.

> in order to comply with Maxwell's equations.

Maxwell's equations are a statistical summarization of the actual behavior.

> It can't work with absorption and re-emission - because the "incoming angle =
> outgoing angle" law of specular reflection is a result of a light wave's
> interference with itself. 

Sort of. What would keep interference from working between an absorbed 
photon and a re-emitted photon?

> But as soon as a photon is absorbed by an electron,
> the photon's probability wave collapses, 

This is incorrect, as far as I understand it. OK, well, for the kind of 
"absorb" you're talking about, where the photon turns into a higher energy 
band of electron "orbit", that might be right. But not for simple "change of 
direction" kind of absorption.

> so the re-emitted light's probability wave has no way of interferencing with it, 

Also incorrect. You can get interference between two photons that aren't 
even in the same light cone any more.

> It also can't work with absorption and re-emission

Perhaps my reference to florescence has made the rest of my statements mean 
something to you different than what I intended. When I say "absorption and 
re-emission", i'm talking at the level of individual photons interacting 
with individual electrons in a QED sort of way. I'm not talking about 
absorption and then re-emission some (theoretically) measurable time later, 
like you get with fluorescence.

> because both are limited to certain wavelengths for many materials, 

I think we're talking about different types of absorption and re-emission. 
I'm talking individual photons interacting with individual electrons. I.e., 
I'm talking about the scale where it's nonsensical to argue whether it's the 
"same" photon or a "different" photon.

> Note however that quite a lot of instances of diffuse reflection are still
> "bouncing" light. 

Light "bounces" off the electrons of an atom. Whether you want to call it 
"absorb and re-emit" or whether you want to call it "bounce" simply depends 
on whether you want to think of it as the same photon or a different photon, 
which is a question that makes no sense.

> You also typically get re-emissions at different (usually longer) wavelengths
> than the absorbed photons (if only because of the principle of entropy).

Yes. I confused you with my mention of fluorescence. Sorry about that.

-- 
   Darren New, San Diego CA, USA (PST)
   The NFL should go international. I'd pay to
   see the Detroit Lions vs the Roman Catholics.

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Darren New <dne### [at] sanrrcom> wrote:
> clipka wrote:
> Of course this is true with flourescence, because you're emitting a
> different frequency than you absorbed. It's not true in general. See, for
> example, a laser, where all re-emitted light is specifically in the same
> direction and phase as the absorbed photon that triggers the re-emission.

Nay. If a laser would work this way, it wouldn't be a lAser ("Light
*Amplification* by stimulated emission of radiation"). The laser effect is a
totally different story.

In a laser, the lasing medium are "pumped" in some way by an external source;
typically, this is a high-energy (i.e. short-wavelength) radiation source
outside the lasing medium, like an UV lamp. *That* is what the electrons in the
lasing medium actually absorb.

The emission then happens in a totally different, lower-energy spectral band,
and can happen spontaneously; or it can be stimulated by light of that very
frequency and will then have the same direction and phase as the stimulating
photon; but the stimulating photon is *not* absorbed.


> Maxwell's equations are a statistical summarization of the actual behavior.

Nay. They are an *exact* description of the *waveforms* which *determine*
statistical behavior (the "probability wave" of a particle between actual
interactions).


> Sort of. What would keep interference from working between an absorbed
> photon and a re-emitted photon?

The fact that the photon has at last stopped being a wave (having just a certain
*probability* to be somewhere) and started being a particle (actually
interacting *somewhere* particular).

It is also described of the "probability wave" of the photon having "collapsed".

That's the moment where Maxwell can go home, and have a break until another
photon is emitted.


> > But as soon as a photon is absorbed by an electron,
> > the photon's probability wave collapses,
>
> This is incorrect, as far as I understand it. OK, well, for the kind of
> "absorb" you're talking about, where the photon turns into a higher energy
> band of electron "orbit", that might be right. But not for simple "change of
> direction" kind of absorption.

What on earth should a "change of direction kind of absorption" be?

Strictly speaking, there is no such thing as a change of direction in a photon
anyway - there is just interference of its probability wave with itself,
because of disturbances in spacetime ("gravitation lenses") or the
electromagnetic field (due to the presence of charged particles). Which, at a
non-relativistic, non-quantum newtonian scale, can be described by the concept
of "light rays", but in quantum world there is no such thing.


> > so the re-emitted light's probability wave has no way of interferencing with it,
>
> Also incorrect. You can get interference between two photons that aren't
> even in the same light cone any more.

I'm not sure about this one - but if this is actually the case, then the effect
is obviously limited to the time when *both* photons "travel", i.e. exhibit
probability wave nature. As soon as one of the photons is absorbed, the other
can't interfere with it anymore.


> I think we're talking about different types of absorption and re-emission.
> I'm talking individual photons interacting with individual electrons. I.e.,
> I'm talking about the scale where it's nonsensical to argue whether it's the
> "same" photon or a "different" photon.

Individual photons can't interact with individual electrons unless they collapse
their probability wave and choose a particular electron to interact with.

At that moment, the interaction of the photon with the electron constitutes a
quite precise "measurement" of the photon's location (which turns out to be
identical to the electron's, with only the wavelength and the electron diameter
posing a bit of uncertainty), and therefore according to Heisenberg's
uncertainty relation the photon's impulse (and therefore direction) gets
somewhat wishy-washy.


> Light "bounces" off the electrons of an atom. Whether you want to call it
> "absorb and re-emit" or whether you want to call it "bounce" simply depends
> on whether you want to think of it as the same photon or a different photon,
> which is a question that makes no sense.

Fine. So we have "bounce" back in the dictionary, which was my initial intention
in this argument.

But the question *does* make sense from a quantum physics point of view: There's
a difference between a photon's *probability wave* "bouncing off" a whole bunch
of electrons all at once (by interferencing with itself due to the disturbance
in the EM field caused by the charged particles), or a photon actually
*interacting* with an individual electron in an "absorb and re-emit" kind of
fashion.

The former doesn't do anything to any individual electron. The latter does.

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Darren New wrote:

> 
> [Snip]
> 

So ... now we're starting to get an idea of how light behaves at a 
quantum level? [Sigh] ... The simple "light bounces randomly" is so much 
easier of an explanation ;)

-- 
~Mike

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Mike Raiford wrote:
> Darren New wrote:
> 
>>
>> [Snip]
>>
> 
> So ... now we're starting to get an idea of how light behaves at a 
> quantum level? [Sigh] ... The simple "light bounces randomly" is so much 
> easier of an explanation ;)

Aye to that. And DKB said, "let there be light," and it was good, 
although only an approximation to what he had in mind. :)

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clipka wrote:
> but the stimulating photon is *not* absorbed.

OK. But what I was saying is that your statement ""If you deal with 
fluorescence you'll see that *all* re-emitted light is *undirected*"" isn't 
true in general of all re-emitted light, since non-spontaneous emission can 
be caused to be directed.

>> Maxwell's equations are a statistical summarization of the actual behavior.
> 
> Nay. They are an *exact* description of the *waveforms* which *determine*
> statistical behavior (the "probability wave" of a particle between actual
> interactions).

I'm pretty sure we're agreeing here. "description of the probability wave 
which determines statistical behavior" is pretty much a "statistical 
summarization of actual behavior (of individual particles)", in my book.

>> Sort of. What would keep interference from working between an absorbed
>> photon and a re-emitted photon?
> 
> The fact that the photon has at last stopped being a wave (having just a certain
> *probability* to be somewhere) and started being a particle (actually
> interacting *somewhere* particular).

But if you don't actually "measure" that interaction, it doesn't turn into a 
particle. You can get interference between particles that have already been 
measured, so I'm not sure what your point is?

I suspect we're both, at this point, arguing nonsense, as it's impossible to 
tell whether a photon has "reflected" off an electron or been "absorbed and 
reemitted" as another photon. You're arguing you can tell the difference. 
Everything I've read says photons "interact" with electrons in exactly one 
way, characterized by electron charge, electron mass, and polarization of 
the respective particles.

> It is also described of the "probability wave" of the photon having "collapsed".
> 
> That's the moment where Maxwell can go home, and have a break until another
> photon is emitted.

How do you measure the difference between a photon being absorbed and an 
identical one being reemitted, or the same photon taking a different path, 
or a photon "reflecting" off an electron?

> What on earth should a "change of direction kind of absorption" be?

Like what happens when light goes through a lens. It interacts with the 
electrons in the glass and winds up moving at a slower speed therefrom. Why 
would a lens change the average speed of light if not because it's spending 
time being absorbed and reemitted?

> Strictly speaking, there is no such thing as a change of direction in a photon
> anyway 

In QED terms, there's no such thing as the direction of a photon at all. :-) 
That's why I said above we're both kind of arguing nonsense now.

>>> so the re-emitted light's probability wave has no way of interferencing with it,
>> Also incorrect. You can get interference between two photons that aren't
>> even in the same light cone any more.
> 
> I'm not sure about this one - but if this is actually the case, then the effect
> is obviously limited to the time when *both* photons "travel", i.e. exhibit
> probability wave nature. As soon as one of the photons is absorbed, the other
> can't interfere with it anymore.

This is counter to my understanding.
http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser
You can wait until after the photon has hit the detector, then determine 
whether you'll see interference or not.

> Individual photons can't interact with individual electrons unless they collapse
> their probability wave and choose a particular electron to interact with.

I don't believe that's correct. I'm certainly not well-schooled in this, but 
I don't believe an "unobserved" interaction with an electron by a photon 
will collapse the probability wave, or one would not have "virtual photons" 
and "virtual electrons" interacting with real photons and electrons.

You may be right, tho.

> But the question *does* make sense from a quantum physics point of view: There's
> a difference between a photon's *probability wave* "bouncing off" a whole bunch
> of electrons all at once (by interferencing with itself due to the disturbance
> in the EM field caused by the charged particles), or a photon actually
> *interacting* with an individual electron in an "absorb and re-emit" kind of
> fashion.

OK.

-- 
   Darren New, San Diego CA, USA (PST)
   The NFL should go international. I'd pay to
   see the Detroit Lions vs the Roman Catholics.

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