Wikipedia:Reference desk/Archives/Science/2020 December 14

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December 14

How small could red green blue camera pixels get?

What shrinking levels are banned by laws of physics and what just need sufficiently advanced technology? Sagittarian Milky Way (talk) 22:55, 14 December 2020 (UTC)

Have a look at the article on quantum dots. This shows what happens when particles of a given material are made so small that their physical properties deviate significantly from those of the bulk material. Bonkeyballs (talk) 00:42, 15 December 2020 (UTC)

The number and size of pixels in digital cameras are strongly linked to Moore's projection that that the number of transistors in a dense integrated circuit (IC) doubles about every two years. 12 Years ago a review of this progress stated that "...it appears that the number of pixels available on a sensor is doubling every ~2 years." The linked article identifies enabling factors and recent trends. Gordon Moore has stated that "[his law] can't continue forever...In terms of size [of transistors] you can see that we're approaching the size of atoms which is a fundamental barrier,..." but the Reference Desk can't give predictions. 84.209.119.241 (talk) 01:03, 15 December 2020 (UTC)

What happens if Moore's law lets you make ones the size of photons? Do they start quantum tunneling a lot when classically they would hit the edges or be unable to fit? Will light photons from the same point start hitting different microns from coming in at different angles from hitting different parts of the lens at f/1.4? Sagittarian Milky Way (talk) 01:25, 15 December 2020 (UTC)

Without special tricks, the smallest feature you can resolve using light of wavelength λ is on the order of λ. See Rayleigh criterion for the simplest case. So even if you could make pixels on the sensor smaller than that, it probably wouldn't make much sense, at least not for cameras as they're currently designed. The pixels would just be noisier, without being able to resolve any more features in the image. --Trovatore (talk) 01:28, 15 December 2020 (UTC)

That's what I suspected, in the traditional non-interferometric camera they would squish in but not help the picture. So if the pixels ever get 4 times smaller (~0.95 microns) the noisy pixel brightness limit would get 16 times brighter if all other things stay equal which doesn't sound too bad, if it's too noisy at ~973 megapixel you can always just make the picture ~60 or 20 MP or less before or after you take it and still have a big photo. Sagittarian Milky Way (talk) 02:14, 15 December 2020 (UTC)

(By the way, that's the wavelength of the light, not the "size of the photons", which as far as anyone knows is zero.) --Trovatore (talk) 01:29, 15 December 2020 (UTC)

Huh, I had assumed that wave-particle duality made it not really wrong to picture fuzzy balls about 1 wavelength wide zipping along at c. Sagittarian Milky Way (talk) 02:14, 15 December 2020 (UTC)

So I should say that this is not really my field of primary expertise. But I'm pretty sure that picture is indeed very very wrong. You have two basic alternatives that are less wrong.

One way is to consider light as primarily a wave, the amplitude of which changes in a jerky way when it interacts with electric charges. The photons represent the jerkiness. This is probably the most fruitful visualization for most purposes.

The other way is to think of photons as size-zero point particles, but take a non-realistic approach to their position. In any given Feynman diagram containing photons, the individual photons are points. But there is no answer to which Feynman diagram is the "true" history of the world; it's a sum over possible histories, with different probability amplitudes. This is the path-integral formulation of quantum mechanics. --Trovatore (talk) 02:25, 15 December 2020 (UTC)

Like sound waves except spheres of constant electromagnetic mumbo-jumbo strength per wavelength-wide shell instead of pressure anomaly energy, but not quite electricity cause there's only variations in the electromagnetic equivalent of the gravitational field and no net electron motion (unless you have a clear live conductor in light of course, but that electromagnetic field has little to do with the light), and not needing a medium or ether of course, and somehow having electric and magnetic waves at right angles, and weird electromagnetic things happen when it hits most things? (though many gasses do not affect it much on the scale of meters) Still too wrong? Sagittarian Milky Way (talk) 02:53, 15 December 2020 (UTC)

Sorry, Sagg, I tried reading that but it defeated me. I don't understand what you mean well enough to address it.

Let me put it another way, by way of a different example. Unlike a photon, a proton has a nonzero radius. A proton can also be thought of as a wave, and that wave has a wavelength, the de Broglie wavelength of the proton, which depends on how fast the proton is moving. That wavelength determines, for example, the way that the proton will diffract and interfere if you use it in the two-slit experiment.

But that wavelength has nothing at all to do with the proton's radius.

Why does a proton have a nonzero radius? Basically because it's not a single particle. It's made up of three quarks. These occupy a probabilistic cloud relative to the center of mass of the proton, and if you look at the region where the quarks are most likely to be found, then that's what we think of as the proton, and it occupies a bit of space.

But a photon, as far as I know anyway, isn't like that. There is uncertainty as to where the photon is located in your laboratory frame. But there's no uncertainty as to where the photon is located relative to the photon. It occupies zero space.

Again, I'm not a super-expert on this stuff; someone may be able to point out something I've misunderstood. But this is my understanding. --Trovatore (talk) 19:53, 15 December 2020 (UTC)

Yes it clearly has to be positional uncertainty which makes 100nm pixels fuzzy cause the only other option would be photon diameter which is wrong. I have a small line of 4 small LEDs that emit mostly in a thin beam cause they're at the bottom of skinny holes and their beams are normal at a few centimeters out, do weird things and then look like a quincunx at a range of feet or yards, is the wave nature of light the cause? It's not bright or foggy enough for me to actually see the beams like a raygun, just the dots like a weak laser pointer. Sagittarian Milky Way (talk) 20:35, 15 December 2020 (UTC)

I'm guessing the aperture is square? The Fourier transform of the interior of a square looks like a cross, and therefore that's what you'd expect to see in the far field. See Fourier optics. --Trovatore (talk) 20:45, 15 December 2020 (UTC)

If the aim of shrinking the pixel size is to get more detail in the final image, bear in mind that there are other ways of achieving this. We already have gigapixel images, of which "this is an example from a wedding".. Mike Turnbull (talk) 11:56, 15 December 2020 (UTC)

A gigapixel sensor would still have niche uses. Sagittarian Milky Way (talk) 14:02, 15 December 2020 (UTC)

Why do so many explosives feature the elements of life i.e. H, C, N, O?

For example, gunpowder (NaNO₃, C, S); nitroglycerine C₃H₅N₃O₉; nitrocellulose (C₆H₉(NO₂)O₅)_n; and ammonium nitrate NH₄NO₃.

Or, put another way, why are many explosives organic, as opposed to inorganic?

Our article on explosives says, "explosives are substances that contain a large amount of energy stored in chemical bonds" which seems to suggest that a lot of energy would required to break these bonds? --- Sandbh (talk) 23:10, 14 December 2020 (UTC)

It's because there are structural elements within those compounds, nitrate groups in particular, that will readily decompose into more stable forms such as gaseous nitrogen N₂, and release a lot of heat when they do so. Organic explosives (e.g. TNT, RDX) tend to be preferred over inorganic explosives (e.g. metal azides) because they can be stored and detonated in a controlled manner, are less shock-sensitive, and are generally more reliable. Bonkeyballs (talk) 00:31, 15 December 2020 (UTC)

Also consider that those elements are very common on Earth. Suppose hypothetically that a useful explosive could be made in replacing the carbon and nitrogen in nitroglycerine with rhodium and palladium; still nobody would be interested in manufacturing it (unless it was a lot better in some way that nitroglycerine) because it would cost too much. The "elements of life", as you put it, are relatively cheap. --174.95.161.129 (talk) 03:00, 15 December 2020 (UTC)

What could gold be used for if it was cheap? Wires near seawater? A better I just want a mass than tungsten for some uses if cheaper? Niche bullets maybe? Or too soft? Unless you add some other dense metal? Farm tractor and keel ballast? Pocket scale weights? Wicket bails for windier days? Really cheap harder-to-check-by-eye loaded dice? Sagittarian Milky Way (talk) 03:42, 15 December 2020 (UTC)

One reason is that the decomposition products are gases. This makes it more likely that elements commonly found in gases will be there. Secondly the substance need to able to bond in different ways, a high energy state and a low energy state. This will favour elements that have more than one bond, and can be arranged in different ways in the compounds. Graeme Bartlett (talk) 10:30, 15 December 2020 (UTC)

• There's two things that go into a chemical reaction and the speed at which it will release energy: thermodynamics and kinetics.

• Thermodynamics is primarily a state function, and deals with the difference in potential energy between the starting materials and end products. In order to explode, you need to release a lot of energy quickly, and thermodynamics deals with the "a lot of energy" part; in order to release a lot of energy, you need to have starting materials that have a large amount of potential energy (unstable bonds and non-zero oxidation states) and end products that have low potential energy (stable bonds and zero oxidation states). Nitrate fits the bill for both of these, as they have highly unstable nitrogen-oxygen bonds that force nitrogen into an extremely unstable +5 oxidation state, and produces dinitrogen as a product, one of the most stable substances known to man. You also need to do something with the oxygen, and that's where the carbon and/or sulfur comes into place. Carbon dioxide has a very stable C=O double bond (two of them in fact) and sulfur forms very stable sulfur-oxygen double bonds as well, as in SO2 and SO3. So, nitrate groups have the "a lot of energy" bit down. The other part is "quickly", which is a real concern. After all, if the process takes 1 billion years to complete, the fact that it releases lots of energy is irrelevant to its explosiveness, it will be almost unnoticeable.
• Chemical kinetics is the study of what goes on "in the middle" of chemical reactions, that is what happens between the starting and ending states. One of its concerns is how atoms and molecules bump into each other to undergo reactions. In this case, you need to consider how do you get lots of collisions to happen correctly and in the right orientation and often. This is where the nature of the shapes of the molecules, the orientation of the atoms, and the particular conditions in which the molecules can bump into each other matter. In each of these cases, the reactions are exothermic enough to be self-sustaining once they begin, AND they have a low activation energy, meaning that they don't take a lot of energy to initiate. This is due to the fact that the oxygen in the nitrate groups is hanging out in space, very easy to access, and has two lone pairs of electrons in easy-to-access orbitals that allow for easy bond formation to electrophiles. What's a good electrophile, you may ask? Well, a carbon bonded itself in an unstable bond to a strongly electron withdrawing group, like a nitrate group. So, in the same molecule, you have 1) a nitrate group with oxygens just itching for form new bonds to electrophiles and 2) electrophilic carbons bonded to those self-same nitrate groups. Two neighboring molecules basically want to react with each other. You couldn't ask for a more kinetically favorable situation. Which is why it is good at exploding.

I hope all of that makes sense. --Jayron32 13:41, 15 December 2020 (UTC)