What proof is there that photons are actually reflecting off matter and not just emitted?

Question

How do we know whether or not light is actually bouncing off everything, versus light excuding out of everything? Maybe there is some sort of signal that comes from a light source that tells matter to emit photons instead of bouncing off?

Maybe it's the inverse of what we generally blieve?

for example an apple is seen as Red because other colors are "absorbed", and red is reflected. So what if it's not reflected, instead it's emitted? It only emits the red because some signal is telling it too, once the signal is cut off, it immediately stops emitting?

but what if a "light source" is actually signalling matter to emit light and once the signal is stopped the emission is stopped?

then maybe there's a way to "artificially" trick objects into emitting light without heat?

Answer 1

Simply. Photone is the particle of very very high activating energy. for a photone to be emited from a source, a large amount of energy (i.e. heat) has to be added to that source, in order to change the energetic state of it's compounds. Energy coming from the Sun is not sufficient to heat the possible sources enough to start emiting photones. In neon-lights the needed energy is provided by electrizing the gas in the tube. In regular light bulb the light comes from overheated tungsten threaded wire, which the electricity goes through.

Only some certain chemicals have ability to emit photones at regualr environment temperature - some molecules that involve phosphorus. Phosphorus emits photones (glows in the dark - luminiscates) in combination with some other elements, but the chemical reaction behind the luminiscence is the one of high released energy, in small amounts, so the firefly can glow in the dark. But there still is a kind of high activating energy.

On the other hand, photones themselves do not contain enough energy to induce light emmiting from the surface of low energy. Einstein's famous experiment with photo-effects has proved that only specially prepared materials can emit electrones, when lighted with the light of a specified wavelength. So the electrones are emited, not photones, because of lower activating energy needed for electrones than for photones.

Only when the light source is highly aimed to one single point, with very high energy provided for photones themselves, as in laser, the heating of the surface could reach the temperature of photone emitting. So, YOur assumptions are incorrect.

That's why the hot Sun glows enough to make daylight, and the cold Moon barely helps You to avoid the large obstacles in Your way at night.

Answer 2

In science, there is the principal of Occam's razor, which “states that the explanation of any phenomenon should make as few assumptions as possible, eliminating those that make no difference in the observable predictions of the explanatory hypothesis or theory.” Your idea, although clever, has too many complications as compared to the accepted explanation. Here are a few.
The refraction index would probably be negative instead of positive, which is a completely different effect than what we see (this is discussed in relation to new materials in the current issue of Scientific American). So it is not just a matter of trading one type of energy for another, there is also the problem of the angle that the light hits the material, which determines the angle that the light would exit the material. Plus, everything that is currently understood of interaction at the quantum level reveals no mechanism to accomplish this feat. For a material to do as you suggest, it would need something along the lines a ballistic super computer to determine the correct exit angle, as well as a mechanism and energy source to create the correct type of particle at the correct level of energy, with the correct wave form and then shoot it in the correct direction and do so in very quick order. On top of this, there is the problem of heat absorption, i.e., a white surface gets less hot than a black surface when exposed to the same amount of full spectrum light. This doesn’t even go into the problems you would run into with semi-transparent objects. In those materials light would need to be emitted out the front and back of the material.

Answer 3

The scattering of light and reflection of light differs from the fact that in the former light is absorbed by the medium and is emitted in various directions.

In reflection the boundary surface does not absorb but simply reflects.

The reflection, refraction and polarization are not different properties; this is explained well taking into consideration the electric vectors of the field. Depending upon the direction of the electric vector (whether they are parallel or perpendicular or in any other direction) a ray is partially reflected, refracted or polarized.

With Maxwell’s equation of field, one can explain how the reflection takes place because of the boundary condition.

Your question is sound and there are answers which involve the whole property of reflection, refraction and polarization.

The answer to your question makes one understand ‘dispersion and absorption of light rays by substances’.

Answer 4

refraction is a different phenomenon from emission
if light is refracted from a surface than it bounces at an angle equal to the angle it hits the surface.
if it is emitted than it disperses in all directions - it becomes a source of light.
a mirror refracts light nearly perfectly = most light is refracted
an apple refracts some wavelenghts but not all = the result is color.

Answer 5

NO MASS! Just momentum, p = hf where h = Planck's constant = 6.626068 × 10-34 m2 kg / s f = the frequency of the photon

Answer 6

You might try checking the frequency, phase, and polarization plane of the light coming back from the object being irradiated. There might be a way to discriminate induced emission from reflection.

Answer 7

because if everything was emitting photons, it would never be dark.