Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I understand there are various colours that light can have. But i was wondering why there is no 'black' light. What is the logical explanation for this? I mean I am expecting an answer that goes beyond mentioning the spectrum details. All 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. But is there any other way?

P.S. I am a programmer and didn't study much Physics beyond high school. This question is not a goof. I am not asking this question for fun. I seriously have this curiosity.

share|cite|improve this question
Light is to black as sound is to silence. Black is the absence of light. – Nic Nov 8 '11 at 12:16
you can destructively interfere light...but that would be a laboratory sort of thing....not practical in real world scenario. – Vineet Menon Nov 8 '11 at 12:26
@Nic: also note that Nic is the absence of anything (because it means "nothing" in Czech). :-) Otherwise this confusing thread is another copy of the antilaser thread here: – Luboš Motl Nov 8 '11 at 13:49
A screen in front of a light source creates a dark shadow behind itself. – Vladimir Kalitvianski Nov 8 '11 at 14:21
Active sonic damping is a reality. Used in headphones and a few other contexts. Doing it is visible light is rather a bigger engineering challenge and requires either a high index of refraction material to delay the light to be damped or a means of predicting the signal. – dmckee Nov 8 '11 at 18:54

12 Answers 12

Such a machine would have to zero out the entropy of the photons it cancels, so it is a form of Maxwell demon. It would have to know exactly which photons it wants to cancel out, and it would have to absorb their entropy and then some. It is impossible to do this at a distance--- you would have to put a black barrier in the path of the photons.

An analogous question: you can make a "heat ray", which heats up objects in its path. Can you make a cold ray which cools everything in its path? Again it is easier to heat than to cool, because cooling requires large entropy production elsewhere, while heating doesn't require anything.

share|cite|improve this answer
You can create a cool jet of gas. – Anixx Aug 5 '12 at 7:13
In an absolute 0 environment that cool gas would actually be a heat ray. – Hobbes Nov 5 '14 at 16:56

Darkness is the absence of visible light. You can't build a machine to output the absence of something, except insofar as it does not output that thing. So a device that just sits there and doesn't produce any visible light at all (like a lamp that is turned off, or most rocks) is as good an example of this machine as you are going to get.

A black hole is something different: it absorbs all light that falls on it, but it doesn't actually produce a thing called darkness.

share|cite|improve this answer
good one, it is obvious, not output any light at all, it outputs darkness (in a dual logic kind of way) :) – Nikos M. Oct 28 '14 at 18:04
maybe the question means, output active darkness, in the sense that it can annihilate lightness (if we can use such term) – Nikos M. Oct 28 '14 at 18:07

This is a very meaningful question in my oppinion. 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. Light follows transport phenomena which is distinguishable and test-ably different from darkness. Could you make a device that propagates darkness in the same way that light propagates? No, you can't.

You mention colors, and I think that in the context of colors your question makes more sense. 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. It would be possible to configure a basis for representing a color where lightness, or conversely, darkness is one of the values involved, sure. But the reality is that colors swaths are not a physical representation of light incident on something in the real world. 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. It is more difficult to think of something that represents darkness in a more straightforward sense with the correct physical model.

All 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. But is there any other way ?

A 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. If light is being pulled in, you are being pulled in as well.

share|cite|improve this answer


Define black.

If black is the usual definition of a colour and the expectation is that it has a frequency and energy h*nu then it exists in the infrared and the ultraviolet regions. Human eyes will see a lamp shining in wavelengths outside its receptors in the retina, as black. One can call it a machine generating black light.

Black defined as darkness means no radiative energy

a machine as powerful as a blackhole; it could bend the light so hard that all we would see is darkness

All that would do is create a vacuum , and yes a vacuum would give off no radiation and human eyes would not see anything: darkness.

share|cite|improve this answer
run out of votes today, will upvote this later – Nikos M. Oct 28 '14 at 18:05

There is a kind of quantum optics experiment known as vacuum squeezing. This is accomplished, at its basic level, by sending a strong laser through a sequence of polarizing beam splitters arranged so that light takes only one path (the other paths cancel out). By introducing non-linear elements such as Rubidium-87 vapor, the path with "no light" can undergo something called squeezing. Simply put, the Heisenberg uncertainty relation between position and momentum of photons in the dark state can appear to be violated (although it isn't if you look at the whole Wigner function).

Is this the sort of answer you're looking for? It's kind of a vague question.

share|cite|improve this answer

If one uses a cloaking device, lights flows around the object cloaked so the cloaked space is pure darkness. There is not yet a cloaking device for the visible light, but it has been done for other frequencies as well as for sound. The cloaked room is completly silent with no sound whatsoever.

share|cite|improve this answer

my understanding of what you are asking is if there can be light with negative energy. The knowledge we have is that, in the current state of our universe that is not possible. 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. 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).

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. We make the source to emit light by closing a circuit, which flows a current, which heats something.

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. 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.

share|cite|improve this answer

Black color is simply a color that does not allow any reflection. Without reflection no visibility could happen. A red color, a perfect red color, absorb everything and reflects only and only red color. While a black color, absorbs every ray which hits it. That is the reason behind its shades also. A dark gray color (with 80% black and 20% white) absorbs 80% of the light it receives and only returns 20% of it. 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.

Now, we have shiny black, which might confuse our answer a little. A shiny black is not a color, but it is a type of material. It is very significant to not miss the point. Assume a sphere which is perfectly black while it also has some reflections. Is it possible? A perfect black to have some reflections? That's it, as I have already mentioned, that is a material. It is a little complicated apparently, but I dare describing it: A color is fully an abstract idea. One of the attributes of matter is color. Then, the constitutive parts of a material which are particle in shape, and express special behavior when contacted by a light ray. They are inherent within the material. Now, a shiny black, like a leather wallet, has two distinct [color-related] materials. A shiny top-coat, and secondary black surface. Shiny material has its own description, but do briefly throw some light upon it, it is a [relatively] noiseless material which does not allow light to distract (diffusion) and allows a focused reflection. It only allows a portion of light to pass throw it (the amount of shininess); and the rest of the rays are reflected from it. Now the amount which pass throw it, reach the black layer. And thus are devoured by black layer. Conclusion: the amount of reflected light from shiny surface is merged (multiplied) by the amount of non-reflected light, and making a black material which has shines over it. So again we see blackness is the absence of reflected natural rays visible by unequipped human eyes.

A last note: Science is separated from metaphysics, it is science that makes decision and should be the criterion of our talks. Metaphysical philosophers were stupid in science. They were lazy to discover the phenomenon, and if they discovered, they probably failed at dissecting it. However, we have scientist who had metaphysical activity; anyway, when we talk about their discovery, we still refer to them as scientist. :)

(I have not corrected the misspellings; sorry for that) The above definition that I have proposed, are from my experience in Animation and Visual Effects.

share|cite|improve this answer

We can never actually emit darkness, its physically impossible from the current knowledge of physics.

To elaborate upon that answer, normally when humans look with their eyes they see the visible spectrum and therefore we cannot see any more or less frequency than that, this therefore results in us either not being able to see it in its true form. Now, consider wearing an infrared camera to our eyes, then we would be able to detect any sort of black-body radiation that is too weak to detect with our eyes.

That said we can now say at temperatures (getting higher), black bodies glow with increasing intensity and colors that range from dull red to blindingly brilliant blue-white as the temperature increases. Or anything that has any temperature glow some black-body

Now, let us assume we do have such a material that can never emit light at all, then using our first statement we can say it must not have any energy of heat in side it, this would mean that the object is literally in absolute zero temperature which is impossible as Heisenberg's uncertainty principle, which says the more precisely we know a particle's speed, the less we know about its position, and vice versa. If you know your atoms are inside your experiment and their respectful masses, there must be some uncertainty in their momentum therefore now if we refer back to the equation of momentum:

$\vec{p} = mv$

we can say since we know the mass of the particles in our experiment, we must conclude it must have some velocity, but since any high-school student will learn temperature is an macroscopic feature of the average velocities of the particles in side it, we must conclude that that it cannot be absolute zero temperature and since we have said previously anything with a temperate environment must also emit black-body radiation. That said we can never reach absolute zero due to quantum mechanical effects that come into play at those temperatures.

As for black-holes, even those are hypothesized to emit very little black-body radiation dubbed hawking radiation after the world-re-known physicist Stephen Hawking, who hypothesized this effect, although not experimentally verified or any observation made pertaining to this quantum effect, its expected to leak our very small energy:

$$ P =\hbar c^2/15360\pi G^2M^2 = 9.004 * 10^{-29} $$

where $M$ which is equal to $1.98855\pm 0.00025 * 10^{30}$ (mass of a solar mass).

Not to mention, black holes are not really an material just an extreme curvature of space-time which results in an lack of light from its event horizon.

Finally, if an object even still managed to get to absolute zero (which is impossible as I have concluded in previous statements) it would then start to violate laws of thermodynamics, therefore its impossible to make such a machine.

This therefore concluded that its impossible to emit darkness (black light) after all black light does not exist but its just darkness

share|cite|improve this answer

There are a couple of things here to clarify.

What is darkness, just (a shade of) black color (as per @anna's answer), yes a machine can do that.

Is it a about a machine that outputs passive darkness (as per @DavidZ's answer), yes it can be done (just dont output anything, it outputs darkness).

Is it about active darkness in the sense that it can annihilate light? Then this would relate directly to a perpetuum mobile of the 2nd kind (the entropy related one) and it would be impossible (as per @Ron's answer)

Note that black holes are supposed to actually radiate (precisely due to entropic/2nd Law considerations).

share|cite|improve this answer

Like The great sir Albert Einstein said to his dumb teacher- " there is no coldness its only the absence of heat ". Analogously we can say that there is nothing such as darkness its just the absence of light . But wait a minute Young's double slit experiment proves that light + light =light and light + light = darkness as well. It's not possible to create darkness producing machine which produces much visible darkness but on a smaller scale YDS appratus can be thought as a darkness producing machine(alternatively bright and dark).

share|cite|improve this answer

It may be worth looking into an alternative theory of colour eg. Goethe. He proposes that colour arises from the interplay of light and darkness. Darkness, then, is not the absence of light but the polar opposite. It is my opinion that human consciousness is not ready to arrive at a true understanding of the phenomenon of darkness. We are still grappling with the mystery of light which remains inextricably linked with the phenomenon of fire and the activities of stars - our sun being the most obvious example. Some people believe that thought and consciousness emit a subtle light. It is the earth's atmosphere that possibly provides the most clues in the search for a better understanding of this topic.

share|cite|improve this answer

protected by Qmechanic Jul 13 '13 at 9:10

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.