It is very significant to not miss the point. A shiny black is not a color, but it is a type of material. Now, we have shiny black, which might confuse our answer a little. While a black color, as you have mentioned, which is a perfect 100% black color, never allows any ray to be reflected off from it. A dark gray color (with 80% black and 20% white) absorbs 80% of the light it receives and only returns 20% of it. That is the reason behind its shades also. While a black color, absorbs every ray which hits it. A red color, a perfect red color, absorb everything and reflects only and only red color. Without reflection no visibility could happen. But there are dynamical reasons as well if we had a source of "negative light" (which is equivalent to a target of positive light), it would be perceived as electromagnetic waves of negative energy, because upon their capture by atoms in the future, they will decrease their energy in order to compensate the energy that was absorbed in the target - Maxwell equations predict a definite positive energy density for electromagnetic waves, so, such solutions are disallowed.īlack color is simply a color that does not allow any reflection. To an extent, is the 2nd law of thermodynamics that "gets in the way" of us making a target to absorb light, since it would allow us to absorb thermal radiation of objects by throwing "black light" into them. We make the source to emit light by closing a circuit, which flows a current, which heats something. This all goes to the subject of causation when we make a source to emit light, we believe the light will be there, propagating in the forward cone, both when we measure it and when we don't like the trees, they will fall regardless if there is someone on the forest. But "black" light would require a photon that will be absorbed from the future cone this we have not ever observed (or at least, we believe we haven't observed them). Photons can be either absorbed from a null ray coming from a past cone, or emitted to a null ray going into the future cone. The knowledge we have is that, in the current state of our universe that is not possible. My understanding of what you are asking is if there can be light with negative energy. If light is being pulled in, you are being pulled in as well. But is there any other way ?Ī black hole always falls short of preventing light from reaching you, with the exceptions of the cases that you are just on the event horizon or below the event horizon. It is more difficult to think of something that represents darkness in a more straightforward sense with the correct physical model.Īll I could think of was a machine as powerful as a blackhole it could bend the light so hard that all we would see is darkness.
For the real world we represent light as a spectrum, where there is a specific intensity for every energy/wavelength in the entire real number line. But the reality is that colors swaths are not a physical representation of light incident on something in the real world. It would be possible to configure a basis for representing a color where lightness, or conversely, darkness is one of the values involved, sure. There are multiple combinations of values that can be used to represent a color, and in computer programs you can see some where a saturation value is used, for instance.
You mention colors, and I think that in the context of colors your question makes more sense. Could you make a device that propagates darkness in the same way that light propagates? No, you can't. Light follows transport phenomena which is distinguishable and test-ably different from darkness. I have heard physicists talk about philosophy of physics classes where they deal with questions like "can light be the absence of darkness"? This might seem like a question lacking concrete implications, but it isn't to at least some extent. This is a very meaningful question in my oppinion.
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