The other side of the debate would be that if the mirror is moving towards or away from the light, the reflection will be Doppler-shifted to a higher or lower frequency. Does this mean that the reflected photons are not the same photons as the incident photons, or does it mean that the same photons have had their energy changed? I think there is no meaningful distinction because every two particles with the same name in quantum mechanics are identical anyway. There's no telling which are which. If I showed you a photon, then took it back and showed you another, you would never be able to tell whether I had opened the same box twice or if I had taken the old one out and captured a new one from my desk lamp.
[1]https://en.m.wikipedia.org/wiki/QED:_The_Strange_Theory_of_L...
>Light can reflect off of transparent crystals
No, the point of the above post is that a photon is not "reflected", but captured and re-transmitted by photo-electric effect.
Light travels at maximum speed C, only when in a vacuum. Otherwise it travels through a medium. We can see that the different speeds of light cause dispersion such as when it enters and leaves a clear prism (changing mediums, thus spreading out the different speeds/wavelengths of light). Since we know that light is "traveling in a medium" when it travels in air or in the prism, then what do we mean by this? It is traveling slower, so it must have some information about the medium...
So how does the light "know" it's in a material, without interacting with the atoms of the material?
Of course, it couldn't know. It is interacting with the material. The behavior of the particle is fully explained by the photo-electric effect.
If light is reflecting off of something, it's precisely because that thing isn't transparent[0].
> if the mirror is moving towards or away from the light, the reflection will be Doppler-shifted to a higher or lower frequency.
This does not refute OP's assertion. If the photon is absorbed and a "different"[1] one emitted, the emitted photon is still emitted from a reference frame that is moving with some velocity, therefore the emitted photon will still be Doppler-shifted... The photon is still emitted at the correct energy level. The photon appears to have higher energy in the stationary reference frame.
[0] Disregarding human-centric definitions that have to do with visible spectrum.
[1] Thinking of the photons as "different" doesn't actually mean much when it comes to quantum mechanics; they're indistinguishable bosons.
> Even in a vacuum you can send a photon in one end and see one on the other and not really know that the one going in was “the same” as the one going out.
I don't really care, if _I or current physicists_ cannot _distinguish_ them, when I ask, whether they _are_ the same. I would rather have an answer like as follows:
"We do not know this yet. We do not have the means of telling, whether it is really the same photon or not. However, even if they were not the same, it would make no difference (according to our current understanding of physics!), as their effect on the surroundings would still be the same, because ..."
Geometrical optics is really sufficient here, it explains it perfectly, and is much more intuitive than other more complex models.
This part is philosophy, not physics. Photons are “indistinguishable particles”. You cannot tell two photons apart. This has a rigorous meaning: if you swap two photons, the state of the universe is unchanged. (This is in contrast to fermions — if you swap two fermions, the state of the universe has its phase shifted 180 degrees.) If photons we’re distinguishable, then swapping two photons would swap them, and the resulting state would be different. The matters from a statistical perspective. Look up Bose-Einstein statistics if you’re interested.
As for whether photons are absorbed and re-emitted or merely reflected, this surely depends on the surface on question.
What about total internal reflection? Why does the light travel through the entire body of a glass prism without issue and then suddenly get absorbed and reemitted at the surface? Glass (and water) are transparent specifically because they don't absorb photons in the visible spectrum, so how is it that they absorb and reemit?
Richard Feynman's "QED" is a great read on this.
It's less hand-wavey. Also, the reflectivity of water is very low at low angles of incidence, rising to 100% at normal angles. Less light reaches the wetted material. https://en.wikipedia.org/wiki/File:Water_reflectivity.jpg
Purple is an artifact of our minds, for example. Does not exist as a distinct wavelength of light.
The object absorbs, does not absorb, emits, whatever wavelengths of light. To us, those have colors, where they are in the visible spectrum.
Take us out of the equation, and?
For example.. if I'm looking at a star, and thinking it's the 'same' photon that came off of that star billions of years ago... that might be true in the vacuum of space, but since it's passed through the medium of the atmosphere, does that mean it's constantly been absorbed/new photons being emitted?
That's how electrons and photons interact. It doesn't make sense to say that they are different photons or the same ones, since you can't observe the interaction.
For anything opaque, you should expect this, because otherwise how could you see something's color if you only saw the same photons as the light source it reflected?
> Before water is spilt, 100% of the light travelling towards that part of the shirt will hit the surface. But now only a fraction of the light moving towards it will hit its surface. This is because the light now has a layer of water to go through. And due to the reflectance of water, not all light at the air-liquid-interface (border between air and water) goes through the water. Some of it is reflected.
This is not clear, or at least needs another step. It is saying that part of the reason the shirt looks darker is because some of the light never had a chance to reflect off the shirt, because it reflected off the water first. But the observer only cares whether light is reflected at all, not whether it reflected off the shirt or the water.
I assume the unwritten part is that this specular reflection is only reflecting light in one direction, instead of diffusely, so if the observer is not in that line of reflection, they won't see the light. But this explanation seems wrong to me, as it implies that a wet shirt will have some brighter highlights at some angles, and yet I have never seen this.
But yes, you're right that shiny things tend to seem darker from most angles, because the observer is usually not in line with most of the reflected light.
In computer graphics, Blinn highlights are a pretty decent approximation for shiny surfaces. (Interestingly, the highlights generally don't take on the color of the object unless it's metallic.)
https://en.wikipedia.org/wiki/Blinn%E2%80%93Phong_reflection...
Car headlights on wet pavement are kind of a worst case. You don't get much benefit from your own lights because the vast majority reflects off at a useless angle, but it does reflect back up at oncoming traffic.
(Impractical startup idea: headlight drones, that fly ahead of your car in rainy weather and illuminate the road in front of you from a more useful angle. Powered by microwaves beamed from the car.)
Reflection on the surface of the water is "free" both ways (from air to air bounce, and from water to water bounce), no energy loss. Travelling through water - similar, lite energy loss, though not exactly zero. The only way we can really lose energy is on the cloth reflection. Therefore (almost) all of the darkness effect should be attributable to increased number of cloth reflections, via the total internal water reflection mechanism.
This also explains why colors are more vivid when the cloth is wet - after all the same ray of white light bounces off the cloth several times, increasing the "color filtering properties" of the cloth.
Apart from index mediation, the water film does something else. For rough/fibrous surfaces, the reflection will be diffuse, i.e. visible from all directions. When a water film is present, the surface becomes smooth, and the reflection will be specular, and only visible in one direction. So in most directions, the material will appear darker.
Conductors are a completely different beast. The reflection off of metals are not solely dictated by the refractive index.
Yes, that's precisely the part I was addressing in my last paragraph. If it's specular reflection, then in "most directions" it will appear darker, as you say, but in one direction it should appear brighter, even shiny. But I've never seen a damp rag be shiny in any direction.
(...and it shouldn't be that hard to see, if that effect is really true. With any other shiny object (polished car, CD, balloon) you see the specular reflection frequently.)
Re-interpretations of their classics, such as "Wanted, Dead or Alive - Schrödinger Redux" failed to capture the live animal energy of the original, and "Livin' on a Fermion", with lyrics such as "Pauli used to work on the docks/That's not even wrong/He'd have fallen right through/It's tough" left not only their core audience baffled, but indeed all sentient beings.
Hardly surprising that Mercury Record buried all mention of the album (and will deny it to this day). They returned to more familiar ground with "Jersey Shore", but lament the death of the Concept Album with this last, brave attempt.
Replace the air gap with liquid water, and the slush will be much darker because the difference in index of refraction at each intereface (now liquid/ice/liquid/ice etc) will be much less.
Basically, in electrical terms, less impedance mismatch so less reflection.
Same thing works for other things besides snow.
But the minute fabric gets wet, it looks so unrealistic. It looks plastic-y or greasy. I can only assume this is an active area of research, but we’re clearly not there yet.
For example, a raytracer can simulate specular (mirror) or diffuse (matte) reflections by simply playing with the incident angle when the ray hits an object. The more jitter, the more matte the object appears. Similarly, you can get very nice "milk" or "skin" coloration by following the incoming ray slightly below the surface before reflecting back out, the same way this discusses having a thin layer of water would prevent some of the light from reflecting back out.
I mean, it obviously doesn't know where they stand.
Snow sunburn is also a thing, IIRC snow can reflect as much as 90% of the UV. You can actually go 'snowblind' from this uv reflection.
On the other hand, when water is added, the above cone of light is broadened, on account of refraction, so it is no longer the case that all the light within it will enter your eye.
These considerations must be modified (I am not sure how) if the water forms a thin, capilliary-adhering film that conforms to the roughness of the substrate.
Finally, if the substrate is transparent or translucent (e.g. sand (quartz crystals)), the change of refractive index at its surface is lessened in the presence of water, causing more transmission / absorbtion and less reflection / scattering there.
Other comments here have put forward the explanation that the water matches the index of refraction of the material much more closely, meaning that light is more likely to pass straight through the material instead of bouncing off. This explanation seems much more likely to be correct to me.
Another experiment: Put some wet spots on a cloth and hold it up to a light source. More light will pass through the wet spots than the dry cloth. This certainly suggests that the reason wet cloth reflects less light is because more is passing through. (I have tried both these experiments with paper towel used as the cloth.)
https://www.quora.com/Why-does-snow-appear-white-in-colour-w...
https://www.quora.com/Why-do-damp-items-such-as-fabrics-appe...
This is not exactly accurate to my understanding. As carl Sagan once said "we are star stuff", and just like stars wherever there is a human to see - even in the "dark" - there is light.
Take the image of the sun/sun rays, moon and eye from the article, its important to remember all matter is emitting light (that includes the moon and the eye and human attached to the eye) not just the Sun.
Even in the "dark" everything including the person will be emitting light (electromagnetic radiation). Humans mostly emit electromagnetic radiation in the infrared wave length, but humans also emit some electromagnetic light in "visible" wave lengths. The issue is obviously humans have not evolved eyes that see "infrared" light like other animals, and similarly the visible light humans emit is below the intensity human eyes evolved to see. However, if human eyes were sensitive enough all humans would appear as shining stars emitting their own light.
Not sure its a thought experiment, but I always thought to understand human sight look at your hand. Then pretend you saw infrared and imagine what you hand looks like (or google a hand in infrared), then do the same as though you saw x-ray wave length light. Now try to imagine if you could see all 3 spectrum at once...what would that look like?
> The issue is obviously humans have not evolved eyes that see "infrared" light like other animals, and similarly the visible light humans emit is below the intensity human eyes evolved to see. However, if human eyes were sensitive enough all humans would appear as shining stars emitting their own light.
In other words, the quote you pulled is completely accurate. It didn't say there was no way to channel the information about the surroundings into any photosensitive device. It says quite clearly that you can't channel the information into a human eye.
> Not sure its a thought experiment, but I always thought to understand human sight look at your hand. Then pretend you saw infrared and imagine what you hand looks like (or google a hand in infrared), then do the same as though you saw x-ray wave length light. Now try to imagine if you could see all 3 spectrum at once...what would that look like?
Probably much like looking at a solid object embedded in colored but not opaque glass. The ability to see things inside other visible things is not foreign to the visible light spectrum.
No it specifically said
>"This is why you can’t see things in dark; no light means...".
Not sure how to make the distinction any simpler, maybe you can follow:
If there is no light, then it is dark (true) - thats what you claim is being said, but thats not what is said, what is said is
If it is dark, then there is no light (false) - any time you have been in the dark, there has always been light.
>Probably much like looking at a solid object embedded in colored but not opaque glass. The ability to see things inside other visible things is not foreign to the visible light spectrum.
Sure the non-imaginative approach is to say just superimpose 3 images on top of one another with each image having some transparency. And sure that may make sense for objects like bone inside the persons outer skin...but emitted heat is not a solid object inside another solid object, it neither embedded in the object (its emitted) nor solid.
https://en.wikipedia.org/wiki/Purkinje_effect
edit: maybe not. Looks like one is physiological, the other physical
In this type of reflection, the angle at which light hits the surface is the same angle at which the light leaves the surface (reflected). This type of reflection tends to happen on smooth surfaces like a mirror or glass.
It feels like the fact that water is absorbed has to matter here--wood, dirt, cloth and sand all darken when they absorb water but not when the water remains in a bubble on the surface (as it does on cloth/sand/wood treated with a hydrophobic coating).
I don't think there's anything wrong with making personal blog posts about learning well established concepts, but promoting them is the reason we have a million Medium/etc articles that are just regurgitating a paper without adding anything of substance.
