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April 7

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Total gear reduction ratio

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I'm trying to work out whether any generalizations can be made about the total gear reduction ratios of various speed-shift gearboxes (aka 'transmissions') used in tracked or half-tracked military vehicles. I'm aware that some people simply use the product of all the ratios, but in my sources the authors multiply the 'jumps' between each ratio (e.g. 9.21 / 4.56 = 2.02 as below), or divide the highest ratio by the lowest (9.21 / 0.69). The result is pretty much the same, I'm not concerned about the exact figures. I think the letters i and q are used in the formulae, but again I'm not too bothered about them.

Thus a certain gearbox is stated to have a total (or overall) reduction ratio of 1:13.4 - the same figure also can be reached by dividing the max. road speed in top gear at max. revs by the speed in 1st gear.

Gear Ratio Jump
     1:x
1 9.21 2.02 2 4.56 1.59 3 2.87 1.56 4 1.83 1.45 5 1.27 1.41 6 0.90 1.31 7 0.69
Overall reduction ratio = 13.4

Using the above methods, here are nine gearboxes in order of overall reduction ratio:

a  8 gears, 11.2 / 0.66  i = 1:17
b  8 gears, 11.0 / 0.68  i = 1:16
c 10 gears, 8.00 / 0.55  i = 1:14.55
d  7 gears, 9.21 / 0.69  i = 1:13.4
e  7 gears, 70.0 / 5.50  i = 1:12.7 (km/h)
f  6 gears  40.0 / 5.5   i = 1:7.27 (km/h)
g  6 gears, 42.0 / 5.8   i = 1:7.24 (km/h)
h  5 gears, 32.4 / 5.1   i = 1:6.35 (km/h)
j  4 gears, 5.58 / 1.00  i = 1:5.58

Letters b and d used the same 23 litre engine, c, f, g & h used a 12 litre engine, and e and j used a 4.2 litre engine. Although a high top speed is desirable, cross-country tractability is probably to be preferred.

Is there any purpose to comparing these ratios? What, if anything, can be deduced from these figures? Cheers, MinorProphet (talk) 12:10, 7 April 2022 (UTC)[reply]

It is unclear what you are asking. The road speed of a vehicle is given as distance per time unit, such as expressed in km/h. The number of revolutions of a gear is a dimensionless number; the rotation speed is given as number per time unit, which has the same dimension as the hertz. You cannot directly derive the gear ratio from a combination of the two, but you can convert distance to number of revolutions of the wheels of the car if you use the wheel diameter. Is there any purpose to comparing the weights of two bullfrogs? Yes, if you are the judge in a competition of "whose bullfrog is the heaviest". Probably not if you are just on a leisurely stroll and happen to spot two bullfrogs. Whether a comparison has a purpose depends on what you want to achieve.  --Lambiam 14:02, 7 April 2022 (UTC)[reply]
Thanks for your reply. What is the purpose of stating the total gear reduction ratio? It's often given in statistics about transmissions, in the way that the power and torque figures of an engine are also given and can be fruitfully compared. Take the gearboxes e and j in the list above. The first is semi-automatic, the other manual. They were both used with the same engine in two prime movers of similar capabilities. What advantage might one have over the other? Cheers, MinorProphet (talk) 19:09, 7 April 2022 (UTC)[reply]
The performance of the combination of engine and transmission depends on many things, not the least of which is what aspects you want to grade this performance on. Is it sheer power? And if so, at which speed? An engineer doing the calculations may need to know the applicable mechanical advantage and thereby the torque ratio. I cannot readily think of a plausible advantage of knowing the gear ratio for a typical car buyer, who should be more interested in the fuel consumption given the envisaged use. I'm not into cars, though.  --Lambiam 07:28, 8 April 2022 (UTC)[reply]
My opening sentence probably wasn't clear enough, but I'm talking [torqueing?] about tanks weighing up to 40 tons, which have somewhat different design criteria. T[h]anks anyway. :) MinorProphet (talk) 20:44, 8 April 2022 (UTC)[reply]
Doesn't this tell you the range of total torque available as output? A vehicle with 1:17 is able to vary the output torque of its engine + gearbox more, compared to 1:5. I suppose for civilians, the clients have a narrower set of use cases, for example optimized for highway-driving, city-driving, hauling, or heavy construction. A tank or military transport might have to meet more varied requirements. It may need overcome gradients and not stall or worse roll down backwards, but still be able to take advantage of level terrains or a lighter loadout for faster traveling. GeorgiaDC (talk) 20:01, 11 April 2022 (UTC)[reply]

Why might a chewing gum brand glue to fingers easier than another?

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One used-up gum can be rolled into a ball and easily pulled off, try that with the other and it always bonds to finger pads strong and never gets pulled off 100% like the other and what didn't get pulled off takes extensive rubbing to fully remove. Sagittarian Milky Way (talk) 17:46, 7 April 2022 (UTC)[reply]

Chewing gum is generally composed of a few different things, usually some kind of resilient plastic substance (called gum base), either an artificial polymer, or natural ones such as latex or chicle, along with sweeteners and flavorings. What in there is sticky? Could be any of them. Some gum bases are going to be stickier than others. Wet sugar is pretty much always sticky; though depending on how long you chewed it and how much you digested out of the gum with your saliva would affect that. Without knowing the specific ingredients in the specific gum, it would be hard to say what is making it sticky. --Jayron32 17:59, 7 April 2022 (UTC)[reply]
See Chewing_gum#Stickiness. Philvoids (talk) 21:14, 7 April 2022 (UTC)[reply]
So, just after you have washed your hands with soap, the wad won't stock to your pads. But when they are dry and greasy it bonds. This effect does not depend on the brand.  --Lambiam 22:40, 7 April 2022 (UTC)[reply]
It absolutely is some composition difference like I dunno polymer length, not skin hydrophobicity. Maybe it'd take longer if I didn't use soapwater to reduce thumb-to-forefinger friction but it still takes a lot of trying to use soap to unfuck residues that have been fucked into thin gluey films from trying to rub the residue off without soap. Sagittarian Milky Way (talk) 00:31, 8 April 2022 (UTC)[reply]
I trust you have done a double-blind experiment to determine that the bonding is brand-dependent. The adhesion with which the goo clings to a surface and the cohesion with which it sticks together are different things, though. Cleaning with (soapy) water only works if the water can creep between the goo and the surface it is clinging to. However, because the goo is hydrophobic, the water is goo-phobic. If it could speak, it would say, uh-uh, I'm not gonna go into that mess.  --Lambiam 07:04, 8 April 2022 (UTC)[reply]
It isn't the-less-sticky-one-was-always-soon-after-obliterating-skin-oil-with-soap-dependant. Both cohesive and adhesive proprieties are different. Yes I know that's why it takes so long, the solvent needs to get between the hydrophobic oleophilic goo and the skin oil it loves so much and that takes a lot of trying to force the contact side to roll over so it can break contact. The water of soaps and detergent should not be completely oleophobic, as letting water dissolve oleophilic things (maybe not all gums) is the whole point of soaps and detergents. Sagittarian Milky Way (talk) 14:42, 8 April 2022 (UTC)[reply]

Double-slit experiment and not wave but only input stream?

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In the experiments of double slits [1], originally "Young's slits" with light which is supposed to have confirmed its wave nature, we now also have flows of electrons, atoms and molecules. Outside I noticed that a flow of water molecules meeting an obstacle, produced waves, the same on the bow of a boat. We therefore have a phenomenon of resonance, which could be found with electrons, atoms and molecules. Would this not be an argument in favor of the corpuscular hypothesis of Newton's light, knowing that a flux can also be considered as a train of impulse with common characteristics with a wave: frequency, period and length of wave ? Malypaet (talk) 23:16, 7 April 2022 (UTC)[reply]

Newton's corpuscular theory argued that the straight lines of reflection demonstrated light's particle nature as only particles could travel in such straight lines. Refraction could be explained if particles of light accelerated laterally upon entering a denser medium. Newton's contemporaries Hooke and Huygens refined the wave viewpoint, showing that if light traveled at different speeds in different media, refraction was easily explained as the medium-dependent propagation of light waves. See Wave–particle duality. Philvoids (talk) 00:53, 8 April 2022 (UTC)[reply]
When I quote Newton's hypothesis, a photon is a simple particle like an electron, an atom, etc... and which has nothing to do with Einstein's. I can then interpret a flux of photons as n elementary rays of photons coming from "Planck resonators" (pulse trains). In this case we are indeed in a fluid mechanics as for water and this flow of photons will then be able to carry a resultant wave created by a disturbance, like the edges of a slit. In this case there is no duality, always flows, sometimes carriers of waves. Has anyone considered this possibility? Malypaet (talk) 22:36, 8 April 2022 (UTC)[reply]
Isaac Newton 1642 - 1727 never used the term "photon" which was coined in 1926 by Gilbert Newton Lewis for what Einstein had called a light quantum. Please see Photon#Historical_development and consider whether you question the awards of Nobel Prizes to Einstein in 1921 (energy quantization explains the photoelectric effect) and to Compton in 1927 (photons have momentum). Philvoids (talk) 13:13, 9 April 2022 (UTC)[reply]
I know that. But today we can also very well use the term photon for Newton's corpuscle, as an alternative hypothesis to Einstein's theory of relativity. In this case for a frequency "ν" we have "ν" photons corpuscles in an elementary flow of corpuscles. Malypaet (talk) 21:20, 10 April 2022 (UTC)[reply]
So, it looks like you're confusing two different things here. The wave-like nature of light is not a vibration of photons. Photons are not the medium of the vibration in the way that, say, water is the medium of vibration for the wake of the boat you describe. Likewise, with matter waves, it is not the vibration of matter; it is that matter is the vibration itself. The Young's slit experiment works for single particles sent one-at-a-time, not just for groups of particles, so it's not like a group of electrons are carrying a wave as a group, like a water wave. And the experiment works in a vacuum, so it's not like the electron (or whatever) is traveling through air and generating a wake, which is might be the source of wave-like behavior. You get interference patterns even if single electrons are sent through one at a time, and even if the experiment is conducted in a pure vacuum. So we're left with two major questions: with what does a single electron interfere with? Itself? Other electrons that will be sent in the future? How does a point particle, which is indivisible, go through two slits at once? These are not simple questions to answer, and if you aren't struggling at a core philosophical level with any answer given to these questions, and don't see inherent paradoxes any answer to these questions give, you aren't doing Quantum mechanics correctly. Neils Bohr supposedly said "Anyone who is not shocked by quantum theory has not understood it.", and Richard Feynman similarly said "I think I can safely say that nobody understands quantum mechanics". What you have there is 1) a scientist who has as good a claim as any to inventing the field of QM, and 2) A scientist who has as good a claim as any to advancing the field further than anyone since its invention (specifically with QED), and they aren't comfortable with the conclusions that QM leads one to. And yet, these are the laws and theories, and the experiments confirm them, no matter how nonsensical they seem, we have a theory that makes predictions that experiments have borne out to shocking accuracy millions of time. --Jayron32 12:10, 8 April 2022 (UTC)[reply]
Even more shocking spooky action at a distance. 80.43.81.32 (talk) 12:35, 8 April 2022 (UTC)[reply]
Even weirder than Young's original experiment are variations such as the Delayed-choice quantum eraser experiment. --Jayron32 12:44, 8 April 2022 (UTC)[reply]
Sorry, but in the original Young's experiment there are 2 halves of an original flow that each aim for their slot, one part goes through it either straight or by hitting the edges (like water on a rock) and the rest is reflected on the outside of the slits. So 2 input streams which can create 2 waves carried by these output streams and which will intersect.
When there is only one particle at a time, you have to be a very bad shooter with a trembling hand not to know which slit will be crossed and then you enter into a problem of probability, which has nothing left to do with the original experience. A photon, an electron, etc... is necessarily smaller than a slit, isn't it?
Figure 2 [1] shocks me for the size of the photon at the entrance to the 2 slits! Malypaet (talk) 22:43, 8 April 2022 (UTC)[reply]
It's not like a stream of water hitting some rocks. It's more like waves in a pool of otherwise stationary water. When the photon is detected somewhere, it has arrived at that exact location and has taken the path leading there. If there's more than one such path, it has taken all of them at the same time. But as long as the photon propagates, it follows every possible path, is everywhere at the same time (although not at the same probability). The single photon goes through the left slit, through the right slit, is reflected in the mirror next to the slit, all at the same time. There's no hitting the edges though. What makes it weird is that it even works with the smallest possible wave, i.e. a single quantum of energy, which can only be detected at one place, still taking every possible path as a wave. The size of a photon is like the size of a wave at sea (and I'm not talking about the height): huge, but it still fits through a small hole a quayside. If you take a laser beam, the length of a photon is something like the inverse of the bandwidth of the laser and the width of the photon is the width of the beam. PiusImpavidus (talk) 10:12, 9 April 2022 (UTC)[reply]
So i can understand that your photon is not an elementary object, but a group object where some are passing by the fisrst slit, some by the second and some reflected between the 2 slit. So , this group is everywhere in the same time ! But for an electron, its size is much smaller than a slit. It cannot aim and precisely cross a slit? Malypaet (talk) 22:09, 9 April 2022 (UTC)[reply]
No, the photon is an elementary object, just like the electron, and it can hit only a single detector at a time. The weird thing is, even when it's a single object, it behaves like a wave going everywhere at the same time. The same is true for electrons. Quantummechanics is weird, much weirder than you imagined. PiusImpavidus (talk) 10:56, 10 April 2022 (UTC)[reply]
An electron has a mass, a supposed size, unlike the photon and if it turns out that the photon has a mass, the wave/particle duality disappears, right? Malypaet (talk) 22:24, 10 April 2022 (UTC)[reply]
An electron has mass, but no size. A photon has supposedly no mass (but we can't really prove that) and no size. Wave/particle duality exists for both. PiusImpavidus (talk) 09:20, 11 April 2022 (UTC)[reply]
Furthermore, when a particle, like a photon or an electron, travels between the emitter and detector (whether in the Young's experiment, or really any path between where it is emitted and where it interacts with some bit of matter (an "observation" in QM-speak)), it takes every available path to get there, even fantastically complicated and strange paths. The chance that it hits any one point of the detector is the weighted average of the probability of each of near-infinite number of possible paths. If you want to really know why Feynman is famous, it's not just for being a charismatic, bohemian personality (though he was that too), it's that he worked out the mathematical shortcuts to calculate the averages between all of those near-infinite number of paths. This was the path integral formulation. --Jayron32 11:53, 11 April 2022 (UTC)[reply]
And the trap penning giving an electron radius ? Malypaet (talk) 20:56, 11 April 2022 (UTC)[reply]
Electrons don't have a known radius. They have an upper bound on their possible radius, given the limits of our ability to measure such a radius. Thus, we can say that an electron is certainly no larger than a certain value. Depending on how one calculates or measures such radius, we get different values. The penning trap method you note only found an upper bound of 10-22m (about 100 yoctometers) however this only means that the experiment shows that the electron is probably not larger than this. It could still have no meaningful radius. More importantly, an electron doesn't need to have a non-zero radius, and in many ways, it is better if it didn't. Since it doesn't need to have one for our theories, and we've never measured a lower bound on its radius, if we treat it like a point particle, it works just fine. --Jayron32 14:51, 12 April 2022 (UTC)[reply]
Malypaet, You are trying to visualise and conceptualise quantum phenomena in terms of the everyday "Classical" macroscopic physics that creates our everyday experiences, and which consequently has molded all our thought processes.
This cannot be done! No way of doing so has been devised in over a century of effort by the best minds on the planet. One just has to accept the logically correct mathematics and consistently demonstrated experimental measurements of the "quantum world" as proof of the existence of aspects of our universe outside the scope of our human sensorium, that do not work in the same way as those aspects within it. {The poster formerly known as 87.81.230.195} 90.200.65.249 (talk) 17:34, 10 April 2022 (UTC)[reply]
"This cannot be done! ... One just has to accept..." is antithetical to dealing with any dogma. All of it. Models, classical, modern or speculative are only as good as the bars or hurdles they have managed to pass over and no better. Times change, people change and often because someone becomes aware of even better models, and sometimes they can even upright the upside down. The education of current thought is paramount, but there are limits to what can be taught. Modocc (talk) 21:48, 10 April 2022 (UTC)[reply]
A theory is valid only as long as it is not invalidated, even if it can take several centuries. You are taking an argument from authority which has nothing scientific about it. Take Planck's formula "E=hν" which you call Einstein's "quantum of light", there you have the energy of a photon which depends on its frequency. Now consider Newton's hypothesis, if "ν" is a number of cycles per second of a wave (hertz), it can also be the number of corpuscles per second of an elementary flux coming from Planck's "resonator". In this case, while keeping a notion of group of objects for the "quantum of light" you find the kinetic energy of a corpuscle (photon of Newton) "E=h=1/2 mc2" and its mass “ mγ=2h/c2”, i.e.:
”mγ=1.474 499 464 763 88 x 10-50 kg “
Value often approached experimentally as by Coulomb's law test [2] and more recently with FRBs [3]. Why is it forbidden to use my classical vision, if it gives simple, coherent results that can be found experimentally, as here for the mass of the photon? Malypaet (talk) 21:50, 10 April 2022 (UTC)[reply]
I am not arguing from authority in favour of (or against) any theory; I am suggesting that it is impossible for any macroscopic (classically immersed) being (such as ourselves) to meaningfully describe quantum phenomena in macroscopic terms, because our minds have no suitable referents. This doesn't mean that we cannot describe such models mathematically, and describe what results from them, just that we can't visualise them. {The poster formerly known as 87.81.230.195} 90.200.65.249 (talk) 22:17, 10 April 2022 (UTC)[reply]
You went further than simply lack of comprehension when you said "...as proof of the existence of aspects of our universe outside the scope of our human sensorium..." emphasis in bold added. None of it is proof. Sometimes the earliest modelers lack enough data to make accurate maps. Modocc (talk) 22:40, 10 April 2022 (UTC)[reply]
Just to be clear, I meant proof that certain things are outside the scope of the human sensorium, (and therefore beyond our ability to conceptualise and describe them in words) not proof of exactly what those things are. However, I do not think we are on the same wavelength, so there is little point in continuing this aspect of the discussion. {The poster formerly known as 87.81.230.195} 90.200.65.249 (talk) 16:09, 11 April 2022 (UTC)[reply]
Newton's classical treatment cannot work as such. Read the history. Modocc (talk) 21:59, 10 April 2022 (UTC)[reply]
I read Newton's story and he describes light well as a flow of corpuscles having mass, with the knowledge of the time. After reading Planck's story on black body radiation, we can very well represent an elementary ray as the shot of corpuscle photons by a machine gun, with such and such a frequency "ν". Over one second you have the power of the shot by adding the kinetic energy of each corpuscle and if you multiply by one second you find the energy corresponding to the frequency, is that clear? Malypaet (talk) 22:50, 10 April 2022 (UTC)[reply]
Yes, but you may have missed my reply 3 days ago regarding the KE and mass that is measured. "See https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass" It differs. Also see https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation#Observations_conflicting_with_Newton's_formula. Photons have twice the expected Newtonian deflection due to gravity, but they also have twice the expected KE for their velocity. Thus you are not wrong nor is it forbidden to take that view, but there is a fair bit of mathematics involved when modeling kinetics and kinematics, one of the most important ones is the relativistic Doppler effect. Modocc (talk) 23:55, 10 April 2022 (UTC)[reply]
you refer me to a formula resulting from an invalidated theory if the photon has a mass, to demonstrate that the photon has no mass. It's a vicious circle and the only solution to this kind of problem is to get out of the circle.
As for the deflection of a mass photon by the gravity of an observed half, in the vicinity of the sun or any other star, I would like to know if the atmosphere or other characteristics due to the proximity would not have they also their share for the other half?
Many other observations can be explained by the mass of the photon/Newton's corpuscle:
Gravity black holes with escape velocity > c.
Dark matter = ocean of photons/mass corpuscles in permanent movement in all directions and bathing the universe, which explains well in spiral galaxies, the orbit of stars around their centers, contrary to the theory of relativity of Einstein.
These photons/particles which leave the visible universe with their mass can also explain the acceleration of the expansion of the universe. Malypaet (talk) 22:48, 11 April 2022 (UTC)[reply]
Note that when a particle and an antiparticle annihilate to create gamma rays it is a system that has rest mass which is conserved throughout. It's a consequence of the principle of mass-energy equivalence. Photons therefore are known to gravitate within Einstein's model which is why the black holes trap them. The deflection is not Newtonian primarily because Newton didn't know that particles' masses changed with reference frames. Similar ideas were put forth for the deflection and for Mercury's orbit, but they didn't standup to scrutiny. Modocc (talk) 00:16, 12 April 2022 (UTC)[reply]
And what about dark matter ? With photon mass it's ok , but with Einstein's model ? Malypaet (talk) 20:50, 12 April 2022 (UTC)[reply]
By particle mass, physicists are referring to a particle's rest mass in its proper rest frame such that its velocity is zero unless they are referring to its relativistic mass. Either way, a photons' mass-energy isn't at rest, hence its simply called energy as it changes with relative velocities (blue or red shifted). Note that given matter-waves the symmetric Doppler relations apply to all masses and energy. Again, photons do tug on stars, but the energy and inertia of the stars is vastly greater than the light they emit. Thus photons are already accounted for by the astronomers' best estimations of their mass-energy. Modocc (talk) 02:05, 13 April 2022 (UTC)[reply]
Everything is relative, because if in a short time you are right, over 13 billion years of light emission whose mass/energy you add up, it is quite different. Malypaet (talk) 22:18, 14 April 2022 (UTC)[reply]
You may want to pick up an older textbook that does a better job of introducing classical and the relativistic Doppler relations. Its the symmetrical Doppler relation that is supported by measurements and must be accurately modeled to have any merit. Modocc (talk) 01:35, 11 April 2022 (UTC)[reply]
If you say that the frequency of light is the number of photons per second, then what is the intensity? What happens if you add two identical light sources together? Experimentally, it doubles the intensity, not the frequency. What happens when you split a beam in two? Experimentally, it halves the intensity, not the frequency. How do you want to explain the photo-electric effect? That was crucial to the development of quantummechanics. How do you create a black-body spectrum, with many frequencies at the same time? No, if the theory uses photons, the frequency must be proportional to the energy per photon and the intensity to the number of photons per second.
If you want to understand physics, you're welcome. If you just want to publish your own theories, write a (soft) sci-fi novel. I might like it. PiusImpavidus (talk) 09:20, 11 April 2022 (UTC)[reply]
That the OP conflated intensity with wave number is unremarkable, especially if the wave packet is a spherical short pulse of light with uniform density. BTW, you wrote " the width of the photon is the width of the beam". No, as a particle it behaves as a point particle with no width that is modeled as part of plane wave that is unsupported by a medium.. Modocc (talk) 12:11, 11 April 2022 (UTC)[reply]
Depends on how you see “width” and the photon. The photon has no width of its own, I agree with that (obviously), but the photon (as a wave) is present over the full width of the beam. The width of the beam is the width of the wave and the photon is the wave. Or the wave is a superposition of photons, but it works for a single photon as well. So I think it's fair to say that the width of a photon is the width of the beam. A photon doesn't even have to be continuous. Did I switch too easily to a wave description? PiusImpavidus (talk) 17:32, 11 April 2022 (UTC)[reply]
It depends. Forty-some years ago one of my teachers naively thought that there is some astronomically small chance that all the air in the room could exit the door and leave us breathless; of course ignoring all the thermal wave interactions that prevent that from happening. In the same vein, the photon is exactly wherever it hits a target. In addition, its interactions with matter's near-fields form the group waves that we are able to measure/calculate with virtual particles as it remains a particle with its invariant power determined by its nonzero Poynting vector. The origin and extent of randomness in those processes like with what the teacher had asserted are model/interpretation dependent and last I checked there are at least seventeen interpretations of the quantum wave functions and they don't even have to be physical. Normally, instead of conflating particles with waves I would just say that particles contribute to waves of different sizes such as the laser beam depending on the particles' characteristics and leave it at that. Modocc (talk) 20:13, 11 April 2022 (UTC)[reply]
I have described the frequency of an elementary flux of corpuscles, whose elementary energy quantum depends on this frequency, which is given by Planck's formula. For the intensity, it's simple, you multiply/parallelize these elementary flows and this will give you one or more beams of greater or lesser intensity, it's still easy to imagine, it's the model of the electric current.
With respect to the blackbody, the higher the thermal radiation energy entering it, the higher the dominant frequency will be, and vice versa. All these light beams of greater or lesser intensity/frequency intersect (there are a lot), have elastic shocks on the material which maintain and contribute to thermal agitation. Malypaet (talk) 21:40, 11 April 2022 (UTC)[reply]
For the photoelectric effect, no change, an elementary flux will eject an electron struck by the number of corpuscles corresponding to the energetic quantum of Planck/Einstein. Malypaet (talk) 21:51, 11 April 2022 (UTC)[reply]
Looking at all the questions so far, it might help if you slow down a bit and get a more clear understanding of the basic components of any theory so as not to become confused by different concepts. Wave nature generally just means an entity has a certain probability of being somewhere at any time. Particle nature generally just means the energy of an entity comes in integer multiples of some ground level. GeorgiaDC (talk) 00:36, 12 April 2022 (UTC)[reply]

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