Those particles have discrete values of mass and electric charge. So it comes naturally that their electromagnetic radiation would come in quantified amounts.Not really.
Not really.In a stable orbital motion, the radius depends on centripetal force. In electromagnetic interaction, it depends on electric charges of the particles involved. Since the charges can only change in multiplication of integer, so does the radius.
In classical physics, there is nothing to stop r1 or r2 taking any value it likes
Since the charges can only change in multiplication of integer, so does the radius.For a given atom, the electron can exist in one of a potentially infinite number of orbitals.
In a stable orbital motion, the radius depends on centripetal force.When a planet is orbiting the Sun in a circular orbit, the attraction between the Sun and planet balances the tendency of the planet to fly off into space in a straight line (hence your comment that "In a stable orbital motion, the radius depends on centripetal force").
In a stable orbital motion,In classical physics, any orbital radius is allowed- as long as the velocity matches it.
I want to know how far we can follow classical physics until it inevitably fails and no reasonable assumptions can be put into it to make it work and agree with observations.Quote from: hamdani yusufSince the charges can only change in multiplication of integer, so does the radius.For a given atom, the electron can exist in one of a potentially infinite number of orbitals.
- The charge on the electron is always -1, regardless of which orbital it is in
- The charge on the nucleus is fixed and positive
- The product of electron charge and nucleus charge is fixed
So how come you can have several different permitted radii, if the product of the charges is fixed?
- That's not it!QuoteIn a stable orbital motion, the radius depends on centripetal force.When a planet is orbiting the Sun in a circular orbit, the attraction between the Sun and planet balances the tendency of the planet to fly off into space in a straight line (hence your comment that "In a stable orbital motion, the radius depends on centripetal force").
- There are an infinite number of combinations of radius and velocity where these forces balance for a circular orbit, so this is not the source of quantization.
- It gets more complicated if you try to account for elliptical orbits of planets, as the gravitational force does not balance the centripetal force for most of the orbit.
- And most planets have elliptical orbits
In atoms, only certain orbitals are permitted (quantization)
- To calculate these orbitals, you need to solve the wave equation for the electron.
- Some of these orbitals are spherical, but others look like a cluster of balloons assembled by a clown. How do you calculate the radius and centripetal motion for these?
If you want a simple understanding, have a look at Bohr's model of the atom, where an electron's angular momentum is quantized (classical physics has no equivalent).
- Or de Broglie's model where the electron has a wavelength, and that wavelength must have an integer number of wavelengths to be stable (classical physics has no equivalent).
- But for a good model, you have to solve the relativistic Schroedinger equation, which gets quite complex for anything bigger than a hydrogen atom. Even a Hydrogen atom is beyond what they are paying me here!
See: https://en.wikipedia.org/wiki/Atomic_orbital#Bohr_atom
Neither side "needs" to be quantised.If we accept that mass and charges are quantized, while k is constant, then it is necessary that value of (k. q. q/m) is quantized.
the radius is a continuous variable.
ω².r³ = k.q1.q2/mI agree that everything on the right side is quantized.
I want to know how far we can follow classical physics until it inevitably failsClassical physics, with no quantisation of orbital energy immediately collapses in an "ultraviolet catastrophe".
Classical physics, with no quantisation of orbital energy immediately collapses in an "ultraviolet catastrophe".There are many versions of classical physics, other than Newtonian and Maxwellian theories. Ultraviolet catastrophe is not the only failure of Maxwellian physics.
The discovery of Planck's constant in the year 1900 was one of the most important discoveries that catalyzed the quantum revolution. What started as a simple idea to resolve one of the greatest physics mysteries of the time, turned out to be the key to unlocking the quantum realm. While Planck assumed that the constant would be 0 when measured the constant had a definite and real value, meaning that there was a lower limit on the universe. Preforming a basic version of this measurement is actually really easy and we explore the process of that measurement in this video using some LEDs and a diffraction grating.
A Simple Method For Measuring Plancks ConstantInteresting, but more or less wrong.QuoteThe discovery of Planck's constant in the year 1900 was one of the most important discoveries that catalyzed the quantum revolution. What started as a simple idea to resolve one of the greatest physics mysteries of the time, turned out to be the key to unlocking the quantum realm. While Planck assumed that the constant would be 0 when measured the constant had a definite and real value, meaning that there was a lower limit on the universe. Preforming a basic version of this measurement is actually really easy and we explore the process of that measurement in this video using some LEDs and a diffraction grating.
We report the high-frequency modulation of individual pixels in 8 × 8 arrays of III-nitride-based micro-pixellated light-emitting diodes, where the pixels within the array range from 14 to 84 μ m in diameter. The peak emission wavelengths of the devices are 370, 405, 450 and 520 nm, respectively.It seems like the experiment doesn't explicitly support the quantization of energy.
Dimensional analysis tells us that energy has mass and length in it. In the equation above, they reside in Planck's constant. So the quantification of energy must come from some things that determine the value of Planck's constant.Let's analyze Planck's formula by plotting it in a graphic. To avoid unnecessary complications, the units are chosen so that every constant has value of 1.
Imagine an electron oscillating up and down in 1 Hz frequency. What's the amplitude of the oscillation which is corresponding to minimum energy transfer, which is h? Why can't we get lower than that while still non zero?Let's approach the problem from different direction, while keeping it simple.
Trying to find analogies between the wave energy confined in a string and matter interacting with light.
0:00 Intro
6:38 Experiments with waves in a string
15:40 Analogies with electron waves
22:50 Changing the standing wave mode in a string using phase
26:49 A hypothetical model for demonstrating quantized wave behavior
32:26 Elastically Inertial Poetry
https://en.m.wikipedia.org/wiki/Rydberg_formulaIt can be interpreted that the amount of radiation energy is proportional to the number of waves, regardless of how much time is elapsed to radiate it. In effect, given the same number of oscillators, low frequency radiation took longer to radiate a specified amount of energy, compared to higher frequency.
As stressed by Niels Bohr,[2] expressing results in terms of wavenumber, not wavelength, was the key to Rydberg's discovery. The fundamental role of wavenumbers was also emphasized by the Rydberg-Ritz combination principle of 1908. The fundamental reason for this lies in quantum mechanics. Light's wavenumber is proportional to frequency
1/L=f/c, and therefore also proportional to light's quantum energy E. Thus,
1/L=E/hc (in this formula the h represents Planck's constant).
So there's no voltage where the light suddenly "turns on".Oh yes there is!
Oh yes there is!
Above that value the intensity of emission increases with current, usually limited by heating which reduces efficiency, but the spectrum is relatively fixed.The video shows that the spectrum is temperature dependent.
Returning to the original question, EMR is not necessarily quantised. Maxwell's propagation laws apply at any and all frequencies, so you can in principle generate radiation at any photon energy you want.Afaik, Maxwell's model for electromagnetic radiation has no concept of photon.
Since the lowest visible energy is about 1.2 eV...If you can see 1330nm.
Afaik, Maxwell's model for electromagnetic radiation has no concept of photon.But ours does. And Maxwell doesn't have a model - he derived a selfpropagating wave from the mathematics of known experimental electromagnetic phenomena.
Where does the photon energy come fromIt's still because of this.
Thermal energy can make a contribution to promoting an electron into an excited state. (which is also why the V/I curve is temperature dependent.)This is sort-of related.
He built a mathematical model based on wave mechanics. In his model, light is propagating electromagnetic wave in a medium.Afaik, Maxwell's model for electromagnetic radiation has no concept of photon.But ours does. And Maxwell doesn't have a model - he derived a selfpropagating wave from the mathematics of known experimental electromagnetic phenomena.
one authoritative source calculates 5 eVDefine "authoritative"...
In his model, light is propagating electromagnetic wave in a medium.No medium was required or specified. μ0 and ε0 are arbitrary constants that relate observation to common units of measurement.
conservation laws suggestYour appeal to the conservation laws is good, but you are applying them the wrong way round.
Returning to the original question, EMR is not necessarily quantised. Maxwell's propagation laws apply at any and all frequencies, so you can in principle generate radiation at any photon energy you want.The reason why we ended up with quantum mechanics was because Maxwell's electromagnetic theory doesn't give accurate predictions against experimental results.
What Planck said was IF you have an ideal particle rattling about in a perfectly elastic box, it can only have discrete energy levels, and you can use this model to predict the UV spectrum etc. If the box does not have defined dimensions then the number of permissible states tends to infinity, hence the black body continuum with a continuous distribution of energy versus wavelength as predicted by the rigid box model for any particular wavelength.
https://en.wikipedia.org/wiki/History_of_quantum_mechanics#Founding_experiments
Founding experiments
Thomas Young's double-slit experiment demonstrating the wave nature of light. (c. 1801)
Henri Becquerel discovers radioactivity. (1896)
J. J. Thomson's cathode ray tube experiments (discovers the electron and its negative charge). (1897)
The study of black-body radiation between 1850 and 1900, which could not be explained without quantum concepts.
The photoelectric effect: Einstein explained this in 1905 (and later received a Nobel prize for it) using the concept of photons, particles of light with quantized energy.
Robert Millikan's oil-drop experiment, which showed that electric charge occurs as quanta (whole units). (1909)
Ernest Rutherford's gold foil experiment disproved the plum pudding model of the atom which suggested that the mass and positive charge of the atom are almost uniformly distributed. This led to the planetary model of the atom (1911).
James Franck and Gustav Hertz's electron collision experiment shows that energy absorption by mercury atoms is quantized. (1914)
Otto Stern and Walther Gerlach conduct the Stern?Gerlach experiment, which demonstrates the quantized nature of particle spin. (1920)
Arthur Compton with Compton scattering experiment (1923)
Clinton Davisson and Lester Germer demonstrate the wave nature of the electron[27] in the electron diffraction experiment. (1927)
Carl David Anderson with the discovery positron (1932), validated Paul Dirac's theoretical prediction of this particle (1928)
Lamb?Retherford experiment discovered Lamb shift (1947), which led to the development of quantum electrodynamics.
Clyde L. Cowan and Frederick Reines confirm the existence of the neutrino in the neutrino experiment. (1955)
Clauss J?nsson's double-slit experiment with electrons. (1961)
The quantum Hall effect, discovered in 1980 by Klaus von Klitzing. The quantized version of the Hall effect has allowed for the definition of a new practical standard for electrical resistance and for an extremely precise independent determination of the fine-structure constant.
The experimental verification of quantum entanglement by John Clauser and Stuart Freedman. (1972)
The Mach?Zehnder interferometer experiment conducted by Paul Kwiat, Harold Wienfurter, Thomas Herzog, Anton Zeilinger, and Mark Kasevich, providing experimental verification of the Elitzur?Vaidman bomb tester, proving interaction-free measurement is possible. (1994)
In atoms, only certain orbitals are permitted (quantization)
- To calculate these orbitals, you need to solve the wave equation for the electron.
- Some of these orbitals are spherical, but others look like a cluster of balloons assembled by a clown. How do you calculate the radius and centripetal motion for these?
If you want a simple understanding, have a look at Bohr's model of the atom, where an electron's angular momentum is quantized (classical physics has no equivalent).
- Or de Broglie's model where the electron has a wavelength, and that wavelength must have an integer number of wavelengths to be stable (classical physics has no equivalent).
- But for a good model, you have to solve the relativistic Schroedinger equation, which gets quite complex for anything bigger than a hydrogen atom. Even a Hydrogen atom is beyond what they are paying me here!
See: https://en.wikipedia.org/wiki/Atomic_orbital#Bohr_atom
The Visible Spectrum of Hydrogen vs. Deuterium
Suzanne Fiacco '01
Abstract:
The purpose of this project is to use a reflection spectrometer to find the differences in wavelengths between the spectrum of hydrogen atoms and the spectrum of deuterium atoms. Hydrogen and deuterium share common characteristics. Deuterium is also known as heavy hydrogen because the weight of deuterium is twice that of hydrogen. Hydrogen is the simplest atom, which consists of one proton and one electron while deuterium is made up of one neutron, one proton, and one electron. Since the physical properties indicate that hydrogen and deuterium are very similar, one would expect their wavelengths to be very similar. In this projects, we calculated three of the visible wavelengths in the hydrogen spectrum to be 656.478 nm, 486.542 nm, and 434.415 nm. For deuterium we calculated that these wavelengths shift to 656.296 nm, 486.409 nm, and 434.295 nm respectively due to the additional mass in the neutron in the nucleus. In doing this experiment we measured the wavelengths in the hydrogen atom to be 656.489 nm, 486.44 nm, and 434.238 nm. For deuterium we measured the visible wavelengths to be 656.295 nm, 486.315 nm, and 434.115 nm. After measuring the intensity verses wavelength for the visible spectrum, we can determine the shift in wavelength for the red, blue, and violet lines as we change the source from hydrogen to deuterium. We conclude that the percent error between the differences in wavelengths between the spectrum of hydrogen and the spectrum of deuterium to be 1.65%, 5.3%, and 2.5% for the wavelengths of red, blue, and violet respectively.
For more information, contact Dr. Catherine Jahncke: cjah@stlawu.edu
Maxwell's equations work perfectly well within their limited scope- they are not meant to be a TOE. I think I have already pointed this out.Don't you want to know what makes it not working in microscopic scale?
What does Maxwellian electromagnetism actually predict in each of those experiments?Maxwell's equations describe the propagation of electromagnetic radiation, nothing more or less.
Actually, slightly less.What does Maxwellian electromagnetism actually predict in each of those experiments?Maxwell's equations describe the propagation of electromagnetic radiation, nothing more or less.
The lift equation F = 0.5ρAv2 tells you how an aircraft will fly, but nothing about tyre skid on landing. So what?
Maxwell's equations describe the propagation of electromagnetic radiation, nothing more or less.I'm not limiting the model to Maxwell's equations. It also includes equations for Coulomb force, Lorentz force, Newton's mechanics, and other non-quantum electromagnetic relationships.
The lift equation F = 0.5ρAv2 tells you how an aircraft will fly, but nothing about tyre skid on landing. So what?
Maxwell's equations work perfectly well within their limited scope- they are not meant to be a TOE. I think I have already pointed this out.Don't you want to know what makes it not working in microscopic scale?
Or rather, what makes it work in most macroscopic scale?
Note that flat earth model and Geocentric model also work perfectly well within their limited scope- they are not meant to be a TOE.
They don't, for example, explain the blue sky.Rayleigh scattering in the atmosphere is a classical continuum effect, entirely consistent with Maxwell.
I'm fairly sure scattering depends on the size of the scattering centres.They don't, for example, explain the blue sky.Rayleigh scattering in the atmosphere is a classical continuum effect, entirely consistent with Maxwell.
Nobody said they do.
Maxwell's equations describe the propagation of electromagnetic radiation, nothing more or less.Well, it's actually "less". They only describe the propagation of radiation through a vacuum. They don't even explain the blue colour of the sky.
I seem to have to keep repeating the bloody obvious: Maxwell's equations describe the propagation of EM radiation,I seem to have to keep pointing out the obvious.
Maxwell's equations only describe the propagation of EM radiation in a vacuum.No. They apply to any medium if you substitute εm and μm for ε0 and μ0. The vacuum value is admittedly an experimental approximation, but none the worse for that.
No. They apply to any medium if you substitute εm and μm for ε0 and μ0.I would more or less agree with that. Let's do the "more" bit first:
To make life easy, we measure and publish dimensionless relative permittivities and permeabilities for various materials (including air and metamaterials) so you can just multiply the vacuum value as appropriate.There are some materials for which we just can't - there isn't a simple scalar relating E and D fields OR the B and H fields.
If you get a high enough field strength practically everything behaves non-linearly.I'm wondering how it would look like under different air pressure.
Even air misbehaves if you try hard enough.
Something like this will give you a clue.If you get a high enough field strength practically everything behaves non-linearly.I'm wondering how it would look like under different air pressure.
Even air misbehaves if you try hard enough.
Do Maxwell equations explain electrostatic and magnetostatic interactions?No, they are derived from experiments that show that a moving charge creates a magnetic field and a varying magnetic field induces a voltage in a conductor. These are essentially dynamic phenomena.
If they don't, what does?
Is it compatible with Maxwell equations?
There are some materials for which we just can't - there isn't a simple scalar relating E and D fields OR the B and H fields.....which is why I said for various materials.
Afaik, electrostatic force follows Coulomb's law. What about magnetostatic force?Do Maxwell equations explain electrostatic and magnetostatic interactions?No, they are derived from experiments that show that a moving charge creates a magnetic field and a varying magnetic field induces a voltage in a conductor. These are essentially dynamic phenomena.
If they don't, what does?
Is it compatible with Maxwell equations?
Hamdani, the biot-savart equation is the counterpart to coulomb, describing the magnetostatic.Can it describe interaction between two permanent magnets?
Can it describe interaction between two permanent magnets?
How about a magnet and a small ferromagnetic material?
To be honest I'm surprised Alancalverd uses a 1/r2 law as an approximation at any range
As outlined by @alancalverd , dipoles don't follow an inverse square law for the field strength they produce. At large enough distances, it's an inverse cube law. (To be honest I'm surprised Alancalverd uses a 1/r2 law as an approximation at any range - but very close it isn't a perfect 1/r3 law, which we both agree on)This article shows the derivation.
The net force acting between the dipole and point entity X will be:
FD = k X x / (R-δ /2)2 - k X x / (R+δ /2)2
we can rewrite the above in the form:
FD = [kXx/R2] / (1-δ /2R)2 - [kXx/R2] / (1+δ /2R)2
For the condition δ <<2R, which was set as one of our assumptions, we are justified to apply the
binomial approximation (1+x)n
≈ 1+nx, or 1/(1+x)n
≈ 1-nx, valid for x<< 1. This reduces:
1/(1-δ /2R)2 to 1+δ /R, and 1/(1+δ /2R)2 to 1-δ /R
The force field equation can therefore be approximated as:
FD ≈ [kXx/R2](1+δ /R) - [kXx/R2](1-δ /R)
FD ≈ [kXx/R2](1+δ /R - 1 + δ /R)
FD ≈ 2kXxδ /R3 or simply FD ~ 1/R3
https://www.gsjournal.net/h/papers_download.php?id=1833
I'm sure you could calculate the precise acceleration of a flat-screen television in a 3T field,Quite possibly zero.
So it experiences a force that depends on the gradient of the B field instead of being directly proportional to B
Classical theories don't tell where the quantization of energy comes from.In fairness this is true.
https://en.wikipedia.org/wiki/Planck_postulate
The Planck postulate (or Planck's postulate), one of the fundamental principles of quantum mechanics, is the postulate that the energy of oscillators in a black body is quantized, and is given by
E = n h ν
where
n is an integer (1, 2, 3, ...),
h is Planck's constant, and
ν (the Greek letter nu, not the Latin letter v) is the frequency of the oscillator.
The postulate was introduced by Max Planck in his derivation of his law of black body radiation in 1900. This assumption allowed Planck to derive a formula for the entire spectrum of the radiation emitted by a black body. Planck was unable to justify this assumption based on classical physics; he considered quantization as being purely a mathematical trick, rather than (as is now known) a fundamental change in the understanding of the world.[1] In other words, Planck then contemplated virtual oscillators.
In 1905, Albert Einstein adapted the Planck postulate to explain the photoelectric effect, but Einstein proposed that the energy of photons themselves was quantized (with photon energy given by the Planck?Einstein relation), and that quantization was not merely a feature of microscopic oscillators. Planck's postulate was further applied to understanding the Compton effect, and was applied by Niels Bohr to explain the emission spectrum of the hydrogen atom and derive the correct value of the Rydberg constant.
Planck's energy equation E = n.h.fThis energy equation is equivalent with power equation, if we take another parameter j as h/t.
I think this is more intuitive,Nobody else seems to.
Does it mean I'm wrong? We haven't heard someone else's opinion about this.I think this is more intuitive,Nobody else seems to.
I think this is more intuitive, for following reason. Suppose we have a radiation source so dim that n=1 and f=1 Hz. Minimum value for E=h Joule. But radiation power is still undetermined. If it's radiated in 1 second then the power is h Watt. If it's radiated in 1000 second, then the power is h milliWatt.In currently more common used form of equation, radiation power is not quantized, due to unrestricted time period. Note that quantization of power equation is mathematically equivalent, but it has no flexibility to change the time period.
In currently more common used form of equation, radiation power is not quantized,And that's fair enough because power isn't quantised in the way that energy is.
Power isn't even a conserved quantity.When power changes, where does the difference go? Does it just appear/disappear? Or Is it merely converted into something else?
When power changes, where does the difference go?You tell me.
There's no adequate justification to extrapolate it to other type of power or energy, such as gravitational potential energy.Do you know that gravity affects photon energy?
Afaik, it changes the energy by affecting the frequency, which doesn't affect the Planck energy equation, nor it's quantization.There's no adequate justification to extrapolate it to other type of power or energy, such as gravitational potential energy.Do you know that gravity affects photon energy?
Afaik, it changes the energy by affecting the frequencyThe statement is meaningless.
Planck's energy equation E = n.h.f,Where does the n come from? E = hf in my universe. But that doesn't mean energy is quantised. Planck's model describes the energy levels available to a particle constrained in a box. Obviously if you have two particles with the same frequency in the box, you have twice as much energy, but the box can be any size you like and a free particle can have any amount of energy you care to give it.
Where does the n come from? E = hf in my universe.Are you living in a different universe?
Planck's Hypothesis
In 1900 Max Planck proposed a formula for the intensity curve which did fit the experimental data quite well. He then set out to find a set of assumptions -- a model -- that would produce his formula. Instead of allowing energy to be continuously distributed among all frequencies, Planck's model required that the energy in the atomic vibrations of frequency f was some integer times a small, minimum, discrete energy,
Emin = hf
where h is a constant, now known as Planck's constant,
h = 6.626176 x 10-34 J s
Planck's proposal, then requires that all the energy in the atomic vibrations with frequency f can be written as
E = n h f
where n in an integer, n = 1, 2, 3, . . . No other values of the energy were allowed. The atomic oscillators could not have energy of (2.73) hf or (5/8) hf.
This idea that something -- the energy in this case -- can have only certain discrete values is called quantization. We say that the energy is quantized. This is referred to as Planck's quantum hypothesis. "Quantum" means how great or of a fixed size.
1 photon -> 1 electron
1 electron -> 1 photon
http://www.phys.ttu.edu/~slee/3301/2018_Fall/F18_3301_Lecture4.pdf
In energy equation E=n.h.f, you can change E with constant f by changing n.Afaik, it changes the energy by affecting the frequencyThe statement is meaningless.
How could it affect one but not the other?
The statement is meaningless.
Where does the n come from? E = hf in my universe. But that doesn't mean energy is quantised. Planck's model describes the energy levels available to a particle constrained in a box. Obviously if you have two particles with the same frequency in the box, you have twice as much energy, but the box can be any size you like and a free particle can have any amount of energy you care to give it.
So haviong establioshed, for the umpteenth time, that energy is not necessarily quantised, we have reduced the question "where does the quantisation come from" tothe same level of pointlessness as "why are unicorns born feet-first?"
Some sources mention that n is the number of photons.Which sources?
A universe containing only 1 photon doesn't allow anyone to exist, let alone to think.Nobody said it did.
Gravitational potential energy and electrostatic potential energy has no frequency in their equations.Nor do they refer to a single entity. You need a "system" of at least two particles for gravitational or electrical potential to exist.
In energy equation E=n.h.f, you can change E with constant f by changing n.Only if there's an interaction with something else.
hf is the energy of a photon. Obviously if you have more photons in a box, you have more energy in the box. That doesn't mean that energy is quantised, only that photons are.Can you add radiation energy by half a photon?
No.hf is the energy of a photon. Obviously if you have more photons in a box, you have more energy in the box. That doesn't mean that energy is quantised, only that photons are.Can you add radiation energy by half a photon?
You could ad a photon with 0.5f to the box, but it would still be a whole photon.It means the radiation energy in particular frequency is quantized. It can be interpreted as photon, but it's not the only possible interpretation. Planck himself didn't interpret it as such.
No. It means that a quantum of radiation has a particular energy, and that energy is directly related to the frequency of the wave that models the propagation of that quantum.The concept of quantum of radiation was created after the discovery of quantization in radiation energy.
That quantization was thought to be necessary to explain the energy distribution of black body radiation.You (Hamdani) have said something like this in a few places.
Planck didn't really suggest all e-m radiation was quantised and it certainly wasn't required to explain Blackbody radiation. Taking an extract from your quotation - he only required that the energy in the atomic vibrations of frequency f was some integer times a small, minimum, discrete energy,. To paraphrase that - all Planck advocated for was quantised oscillators in the walls.
In 1900, the British physicist Lord Rayleigh derived the λ−4 dependence of the Rayleigh?Jeans law based on classical physical arguments, relying upon the equipartition theorem. This law predicted an energy output that diverges towards infinity as wavelength approaches zero (as frequency tends to infinity). Measurements of the spectral emission of actual black bodies revealed that the emission agreed with Rayleigh's calculation at low frequencies but diverged at high frequencies; reaching a maximum and then falling with frequency, so the total energy emitted is finite. Rayleigh recognized the unphysical behavior of his formula at high frequencies and introduced an ad hoc cutoff to correct it, but experimentalists found that his cutoff did not agree with data.[1][3] Hendrik Lorentz also presented a derivation of the wavelength dependence in 1903. More complete derivations, which included the proportionality constant, were presented in 1905 by Rayleigh and Sir James Jeans and independently by Albert Einstein.[3] Rayleigh believed that this discrepancy could be resolved by the equipartition theorem failing to be valid for high-frequency vibrations, while Jeans argued that the underlying cause was matter and luminiferous aether not being in thermal equilibrium.[3]Planck's quantization was an ad hoc correction to Rayleigh's model which fortunately fit with experimental data. His argument was that it takes more energy to produce higher frequency radiation, which provides the cut off he needs to explain black body radiation curve.
https://en.m.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law
The concept of quantum of radiation was created after the discovery of quantization in radiation energy.That's linguistically unlikely.
So Planck wanted quantised oscillators in his cavity walls only. Radiation in general or that exists elsewhere in the world (not inside a cavity at thermal equilibrium with cavity walls) was under no obligation to be quantised.That's impossible.
All he needed was quantised oscillators in the walls, the quantisation of any radiation that could be found inside the cavity would follow automatically. This has some relevance for suggesting one reason why all bits of light you might find anywhere (e.g. not just inside a cavity) would still be quantised: If every bit of light (anywhere) has it origin ultimately in a transition that happened in an atom, then the source of all e-m radiation is suitably quantised, so all the radiation you will find will automatically be suitably quantised.Some e-m radiations come from molecular activities. Some others come from macroscopic mechanical vibration.
Some e-m radiations come from molecular activities. Some others come from macroscopic mechanical vibration.Has em radiation from a macroscopic vibration ever been detected?
Yes, you can derive the blackbody spectrum by assuming radiation is quantised but you just don't have to and historically it does not look like Planck made that assumption. All he needed was quantised oscillators in the walls, the quantisation of any radiation that could be found inside the cavity would follow automatically.
Some e-m radiations come from molecular activities. Some others come from macroscopic mechanical vibration.As @Bored chemist started to suggest - that is not actually obvious. While we can model radio waves as being something that might be produced in an aerial just by charges being accelerated up and down the aerial, it is still almost certainly just a model. Each individual photon (or bit of radio wave) probably is being produced by some interaction(s) within atoms (or molecules with hundreds of interacting atoms) i.e. as a consequence of some fundamental atomic scale production mechanism.
Something like this.Some e-m radiations come from molecular activities. Some others come from macroscopic mechanical vibration.Has em radiation from a macroscopic vibration ever been detected?
However, what we now assume and may use in a modern derivation for the Blackbody spectrum does NOT change what Planck originally assumed and how he started to explain the spectrum.I wonder how can anyone disagree with this.
Something like this.I see no photons.
I wonder how can anyone disagree with this.i.d.k. Do some historical research, find some document showing Planck actually proceeded in some other way initially.
but not every em spectrum is necessarily quantised.OK, once you have quantised energy states then conservation explains why they gain or lose energy in quanta.
Why does a hydrogen atom have quantised energy?It's all down to Heisenberg and Schrodinger. For other atoms, you also need a teaspoon of Pauli.
No.Why does a hydrogen atom have quantised energy?It's all down to Heisenberg and Schrodinger.
I smell a philosopher! Indeed, only God can answer the question "why", and as there is no God, the question is meaningless. But if you want to know "how", I refer to the three hon gents I mentioned earlier.No.Why does a hydrogen atom have quantised energy?It's all down to Heisenberg and Schrodinger.
Grass was green before either of them was born.
I thought it was something to do with the molecular orbitalsYes.
And we model the molecular orbitals as combinations of atomic orbitals.Which would produce some very odd-shaped molecules indeed. Whatever happened to delocalisation since 1963?
So we all look forward to your explanation without using anything that Heisenberg, Schrodinger or Pauli wroteI already gave it.
Whatever happened to delocalisation since 1963?I presume that among the things that happened was that you forgot how it works.
I found at least one statement that was made which isn't necessarily true. In the second video at about 7:00 to 7:30 the narrator discusses the use of a cuboidal shaped cavity for the mathematical model (which is the typical choice). There is a fairly casual statement made that any shaped cavity could have been used, the result would be the same but the maths would just be harder. That is not at all obvious and I don't think it should be stated as established fact.Imagine that you wanted to model a BB of some other shape.
Incidentally, hot hydrogen atoms do have a strong absorption for red light.How hot or cold does it take to produce absorption spectrum?
It's the hydrogen alpha line looked at from the other point of view.
In particular, there are some extra assumptions that seem essential and must be added to the mathematical model to correctly predict that very small and peculiarly shaped cavities will still ultimately produce the usual BB spectrum.Can you specify what are they?
I smell a philosopher! Indeed, only God can answer the question "why", and as there is no God, the question is meaningless. But if you want to know "how", I refer to the three hon gents I mentioned earlier.Learning is basically information collection, filtration, and compression process.
How strongly do you want it to absorb?Incidentally, hot hydrogen atoms do have a strong absorption for red light.How hot or cold does it take to produce absorption spectrum?
It's the hydrogen alpha line looked at from the other point of view.
How strongly do you want it to absorb?Strong enough to be detected by common phone camera unambiguously.
Hot enough to significantly populate the first excited state.
How come you don't know that?
Instead of absorption, the pictures look more like scattering effect.Incidentally, hot hydrogen atoms do have a strong absorption for red light.How hot or cold does it take to produce absorption spectrum?
It's the hydrogen alpha line looked at from the other point of view.
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.4college.co.uk%2Fas%2Fel%2Fhow.gif&hash=5a8aff28e0cfb2dad7c616eb9d3374b6)
http://www.4college.co.uk/as/el/how.gif
(https://www.daviddarling.info/images/types_of_spectra.jpg)
https://www.daviddarling.info/images/types_of_spectra.jpg
(https://sites.ualberta.ca/~pogosyan/teaching/ASTRO_122/lect6/figure05-14.jpg)
https://sites.ualberta.ca/~pogosyan/teaching/ASTRO_122/lect6/figure05-14.jpg
You have not understood the pictures.How strongly do you want it to absorb?Strong enough to be detected by common phone camera unambiguously.
Hot enough to significantly populate the first excited state.
How come you don't know that?
Can it be done under room temperature?
Your statement suggests that there's a minimum temperature limit to show the absorption spectrum, contrary to the pictures which suggest that there's a maximum temperature limit instead. What's your reference?
Your statement suggests that there's a minimum temperature limit to show the absorption spectrumAs I already pointed out.
How strongly do you want it to absorb?And you failed to answer it. Why is that?
You could make it out of small cube shaped BB (for which the maths is known)The general idea of building a new (bigger) shape by joining some shapes which you already know will produce a BB spectrum together (face to face etc.) is reasonably sound.
And you can then rescale the problem - effectively using a different unit of length-Yes, there are some places in the usual mathematical method for deriving the BB spectrum formula where scale is fairly arbitrary. The main problem is just that you could be magnifying the errors in the approximations that are made in that method.
Can you specify what they are? (extra assumptions for small cavities etc.)One of the things you'd like to do is to reduce the discrete nature of the frequencies for the supported modes and make them more spread out or continuous immediately.
And you failed to answer it. Why is that?It looks like you fail to find my answer.
Strong enough to be detected by common phone camera unambiguously.
This tells you about flame absorption spectra.The sample shown in the article is Pb. Turning it into gas requires high temperature. If the element is already gaseous in room temperature, like hydrogen, the flame doesn't seem to be necessary.
https://www.agilent.com/en/support/atomic-spectroscopy/atomic-absorption/flame-atomic-absorption-instruments/how-does-aas-work-aas-faqs
Specifically, it seems that Nature does NOT see the wall of a cavity as a sharply defined hard wall. For example, not all of the radiation is stopped or bounced back at the inner edge of the wall, radiation of some frequencies may penetrate the wall slightly before being absorbed or bounced back etc. Indeed that does happen and is verifiable in other experiments - e.g. X rays can penetrate most materials to some depth.Also, if the wall is not a good electric conductor, radio wave can penetrate it. I discussed the limitations of photon model in explaining electromagnetic radiation in another thread.
Opops!And you failed to answer it. Why is that?It looks like you fail to find my answer.Strong enough to be detected by common phone camera unambiguously.
Heating has three effects.This tells you about flame absorption spectra.The sample shown in the article is Pb. Turning it into gas requires high temperature. If the element is already gaseous in room temperature, like hydrogen, the flame doesn't seem to be necessary.
https://www.agilent.com/en/support/atomic-spectroscopy/atomic-absorption/flame-atomic-absorption-instruments/how-does-aas-work-aas-faqs
The flame is not shown in the pictures I posted either.
Instead of absorption, the pictures look more like scattering effect.It's unfortunate that the diagram doesn't mention temperature requirements for the depicted phenomena to be observed. Although the spectrum looks like Balmer series, which implies that the gas is hydrogen.
The video shows em radiation caused by macroscopic motion.Something like this.I see no photons.
Before I saw this diagram, I wondered where does the absorbed energy go. I guessed that some are transformed into heat, which is then dissipated to the environment.Instead of absorption, the pictures look more like scattering effect.It's unfortunate that the diagram doesn't mention temperature requirements for the depicted phenomena to be observed. Although the spectrum looks like Balmer series, which implies that the gas is hydrogen.
(https://sites.ualberta.ca/~pogosyan/teaching/ASTRO_122/lect6/figure05-14.jpg)
Instead of absorption, the pictures look more like scattering effect.There's some similarity but...
Before I saw this diagram, I wondered where does the absorbed energy go. I guessed that some are transformed into heat, which is then dissipated to the environment.Some is.
The second one is quantization of atomic radiation frequency, which is observed in spectral line emission. Balmer discovered empirical formula to describe the spectral line emissions of the hydrogen atom. Bohr interpreted it as the evidence for the existence of atomic orbitals.
In atomic physics, the Bohr model or Rutherford?Bohr model of the atom, presented by Niels Bohr and Ernest Rutherford in 1913, consists of a small, dense nucleus surrounded by orbiting electrons. It is analogous to the structure of the Solar System, but with attraction provided by electrostatic force rather than gravity. In the history of atomic physics, it followed, and ultimately replaced, several earlier models, including Joseph Larmor's solar system model (1897), Jean Perrin's model (1901),[2] the cubical model (1902), Hantaro Nagaoka's Saturnian model (1904), the plum pudding model (1904), Arthur Haas's quantum model (1910), the Rutherford model (1911), and John William Nicholson's nuclear quantum model (1912). The improvement over the 1911 Rutherford model mainly concerned the new quantum mechanical interpretation introduced by Haas and Nicholson, but forsaking any attempt to explain radiation according to classical physics.Bohr's model was proposed because then classical physicists thought that accelerating electrons must radiate em wave. But experiments with superconductor shows that electric current can flow in circular motion without radiating away its energy. It's generally accepted that superconductivity is created by forming Cooper pairs.
https://en.m.wikipedia.org/wiki/Bohr_model
Bohr's model was proposed because then classical physicists thought that accelerating electrons must radiate em wave.Bohr's model was proposed in spite of classical physics saying that accelerating electrons must radiate em wave.
But experiments with superconductor shows that electric current can flow in circular motion without radiating away its energy.You don't need superconductors to run into that problem.
You don't need superconductors to run into that problem.In superconductor, we can observe the electric current making a macroscopic trajectory.
An ordinary magnet should be impossible for much the same reason.
This sun sparkling water looks like a quantization of sunlight,But of course it isn't really quantized.
This sun sparkling water looks like a quantization of sunlight, although each point should be much brighter than a single photon.You can model the effect as multiple samples of a continuum. No evidence of or requirement for quantisation in the macroscopic observation. As the water waves are moving smoothly, any apparent reflection can also appear to move smoothly - there are no discrete vectors or forbidden transitions.
Different sun position.Do you mean reflection angle?
"Point sparkles" obviously require a point source (or a rare event where the wavelets are all substantially concave) and the source must be behind the observer.Seen from the surface of the earth, the sun is not like a point source. It has a significant angular diameter. But somehow the sparkles seem like being quantized.
Diffuse source, or source in front of the observer, produces diffuse reflections in most cases.
Compared with the angle subtended at the observer's eye by the ocean, it is a pointWhat do you mean? Is the angle close to zero?
https://en.wikipedia.org/wiki/Subtended_angle
In geometry, an angle is subtended by an arc, line segment or any other section of a curve when its two rays pass through the endpoints of that arc, line segment or curve section. Conversely, the arc, line segment or curve section confined within the rays of an angle is regarded as the corresponding subtension of that angle. It is also sometimes said that an arc is intercepted or enclosed by that angle.
The precise meaning varies with context. For example, one may speak of the angle subtended by an arc of a circle when the angle's vertex is the centre of the circle.
But somehow the sparkles seem like being quantized.This is clearly not an example of quantization, perhaps you do not know what is meant by the term quantization?
What do you mean? Is the angle close to zero?The diameter of the sun is about 700,000 km, at a distance of about 140,000,000 km so it subtends an angle of about 0.3 degrees at the observer's eye. At 2 m height, the ocean horizon is about 5,000 m so it subtends an angle of about 89 degrees. That is the difference between "almost a point" and "almost a plane".
Let's compare that to the moon.What do you mean? Is the angle close to zero?The diameter of the sun is about 700,000 km, at a distance of about 140,000,000 km so it subtends an angle of about 0.3 degrees at the observer's eye. At 2 m height, the ocean horizon is about 5,000 m so it subtends an angle of about 89 degrees. That is the difference between "almost a point" and "almost a plane".
You also need to consider the texture of the reflecting surface.Right. To produce sparkling, the water surface must not be flat. It can be modeled as combination of many convex and concave mirrors with various curvature. Those convex mirrors produce smaller images of the reflected object compared to flat mirrors. For observer further away than the focal point, concave mirrors also produce smaller images.
But these are not enough to explain the sparkling effect.I think it perfectly explains the effect.
We know for sure that electronic sensors are spatially quantized, although it isn't necessarily true that every pixel has the same size or sensitivity.But these are not enough to explain the sparkling effect.I think it perfectly explains the effect.
Do you now realize that the sparkling is not quantized?
The third image in #153 shows both "sparkles" and "general glow", so nothing to do with the resolution of the imaging system.It's visibly zoomed in. Which makes an area of the moon surface will fall into larger area on the sensor. This reduces the sparkling effect.
Not at the level shown in your images. It was sometimes difficult to distinguish individual leaves on a whole tree photographed with a 56k camera but the blobs in your images are a lot bigger, and in any case the question is about quantisation of the radiation, not the image receptor!The radiation power follows inverse square law of distance. And the effect from quantization of distance is insignificant in this case.
We know for sure that electronic sensors are spatially quantized, although it isn't necessarily true that every pixel has the same size or sensitivity.I think that the sparkling of water is not quantized. I also think that once you get an idea in your head no amount of evidence or logic can convince you otherwise.
Do you think it has no observable effect?
The same applies to you too. You're already convinced that texture of reflecting surface alone is enough to produce sparkling effects as shown in the videos, which I've also observed myself with my own eyes.We know for sure that electronic sensors are spatially quantized, although it isn't necessarily true that every pixel has the same size or sensitivity.I think that the sparkling of water is not quantized. I also think that once you get an idea in your head no amount of evidence or logic can convince you otherwise.
Do you think it has no observable effect?
The same applies to you too.And everyone else. You seem to be the only one that thinks there is some sort of magical thing going on when it is pretty obvious we are just looking at reflections off of the water.
My hypothesis can be falsified easily by showing the same reflection of the sun by rippling water surface using cameras with different resolution, or pin hole cameras with different sizes.The same applies to you too.And everyone else. You seem to be the only one that thinks there is some sort of magical thing going on when it is pretty obvious we are just looking at reflections off of the water.
So are you going to acknowledge that the 'sparkling water' is not an example of quantization? I think there is a possibility that if we pursue this for a couple of posts it could be resolved.
This sun sparkling water looks like a quantization of sunlight,In what way?
The sparkles look bright and then suddenly disappear instead of fade away.This sun sparkling water looks like a quantization of sunlight,In what way?
Are you saying the intensity is quantised or the wavelength or what?
I guess you can call a video frame rate some sort of "quantisation", but it's a weird use of the idea.The sparkles look bright and then suddenly disappear instead of fade away.This sun sparkling water looks like a quantization of sunlight,In what way?
Are you saying the intensity is quantised or the wavelength or what?
I guess you can call a video frame rate some sort of "quantisation", but it's a weird use of the idea.Do you have a better terminology?
Quantisation, sampling, digitisation and scintillation are not the same thing.
Quote from: Bored chemist on Yesterday at 11:33:51Read the second sentence of reply 173, which is explicit.
I guess you can call a video frame rate some sort of "quantisation", but it's a weird use of the idea.
Do you have a better terminology?
Read the second sentence of reply 173, which is explicit.Do you mean it's the same phenomenon as scintillation?
The effects from pixel size of the sensor is yet to be determined.I just recorded a video of reflected sunlight on a rippling water surface. It looks sparkling when seen using naked eyes. But my phone camera doesn't show that sparkling, either from further or closer distance. Approximately 3 meters vs 1 meters from water surface. My main camera has 50 MP resolution.
Let's make a list of things that affect the sparkle.I also think that the sparkling effect involves saturation of sensor pixels. Which is why dimmer light sources don't produce sparkling.
Brightness of the light source.
Contrast with the background.
Visual size of the light source.
Size of the ripple on reflecting surface.
Angle of reflection doesn't seem to affect much, since the sparkling effect can be seen in both low and high angles.
The effects from pixel size of the sensor is yet to be determined.
This is so tedious, I'm out.You are free to jump in or out of the discussion here, anytime you like. You don't need anyone's permission.
Yes.I guess you can call a video frame rate some sort of "quantisation", but it's a weird use of the idea.Do you have a better terminology?
It seems like you have never starred at sparkling water surface.Yes.I guess you can call a video frame rate some sort of "quantisation", but it's a weird use of the idea.Do you have a better terminology?
Frame rate.
Reply 180 at least moves towards a definition of sparkling, though whether the visual effect is due to saturation of the sensor, scatter in the sensor, or flare in the lens assembly, is not clear.Some sparkling bright spots don't show any flare.
It seems like you have never starred at sparkling water surface.It may seem that way to you. But there's no evidence for your view.
https://youtube.com/shorts/QCE9-DU0xDg?feature=sharedYes.I guess you can call a video frame rate some sort of "quantisation", but it's a weird use of the idea.Do you have a better terminology?
Frame rate.
Some sparkling dots in the short video last much longer than a frame duration.Which is hardly surprising since each originates from a relatively slowly-changing source.
Do you understand detector saturation?Yes. That's what you should say, instead of scan rate.
I didn't say scan rate.Do you understand detector saturation?Yes. That's what you should say, instead of scan rate.
Ok, you said frame rate.I didn't say scan rate.Do you understand detector saturation?Yes. That's what you should say, instead of scan rate.
Perhaps you should "cut to the chase" and tell us how sparkles on the water would look without "quantisation".
Is the attosecond laser quantized?Yes.
The 2023 Nobel Prize for Physics was awarded to a fantastic trio working towards imaging electrons on the attosecond scale. I am an optical physicist, so I naturally want to take a deeper look at why this discovery is important, how it actually works, and what it unlocks for our understanding of the universe around us and how it actually behaves.The video shows that there are multiple photons used, although it's not clear how many.
0:00 Electrons and the world of the minute.
1:22 "Everything in physics starts with Einstein" - Isaac Newton
2:26 Breaking the 6 femtosecond record
5:27 How to build the world's fastest laser pulses
7:05 Ad read
7:50 How to see an Electron
9:03 Why don't you just use a single photon?
What's the real frequency?Badly defined because of the uncertainty principle.
Can a single photon have more than one frequency?What's the real frequency?Badly defined because of the uncertainty principle.
Yes, because of the uncertainty principle.Can a single photon have more than one frequency?What's the real frequency?Badly defined because of the uncertainty principle.
No. E = hν
Yes, because of the uncertainty principle.I don't think that both answers can be correct.
But very easy to determine what it was before you measured it!How is it?
Thanks to indeterminacy you can't predict the energy of a given photon, nor the time at which it will arrive at a point, with absolute accuracy, but as far as the photon is concerned it only has one energy.Having one energy doesn't necessarily mean having one frequency. Mathematically, a single narrow frequency requires extremely long duration of sinusoidal wave.
This video shows the sparkling effect using a relatively low resolution camera sensor.We know for sure that electronic sensors are spatially quantized, although it isn't necessarily true that every pixel has the same size or sensitivity.But these are not enough to explain the sparkling effect.I think it perfectly explains the effect.
Do you now realize that the sparkling is not quantized?
Do you think it has no observable effect?
The presence of other effects, like diffraction, flaring, scattering, and various sizes of the ripple makes it hard to exclusively observe the sparkling effect.In modern digital cameras, including those in smart phones, software filters and image processing algorithms can also affect the results of sparkling effect.
If you are still convinced that there's no quantization effect whatsoever in the sparkling of rippling water surface when reflecting sunlight, I'm curious what does it take to make you change your mind?Evidence.
Quote from: alancalverd on 23/10/2023 15:32:10E = hν
But very easy to determine what it was before you measured it!
How is it?
Use a calorimeter or photocell to determine E.I'm sure I heard about someone doing that.
I've recorded for a new video showing this sparkling effect. In one of the cut scenes, there's visible effect of diffraction caused by trace of wiped sweat on the camera glass. But the sparkling tiny bright spots don't seem to be diffracted.Here are some of the screen shots.
Evidence.What kind of evidence do you need?
But you probably need to start by saying what you think quantization means.I asked Google Bard. I think it gives pretty good answer.
Quantization is the process of converting a continuous signal into a discrete signal. This is done by mapping the continuous signal to a finite set of values. Quantization is used in a variety of applications, including digital signal processing, image processing, and machine learning.
There are two main types of quantization:
Uniform quantization: This is the simplest type of quantization, where the continuous signal is divided into equal intervals and each interval is mapped to a unique value.
Non-uniform quantization: This type of quantization is used when the continuous signal has a non-uniform distribution. In this case, the intervals are sized differently to better represent the distribution of the signal.
Quantization can be a lossy process, meaning that some information is lost in the conversion from a continuous signal to a discrete signal. However, the amount of information lost can be controlled by the number of quantization levels used. More quantization levels will result in less information loss, but will also require more storage space.
Here are some examples of quantization:
When you take a digital photo, the camera quantizes the continuous light signal into a discrete set of pixel values.
When you encode an audio signal for MP3 playback, the audio signal is quantized to reduce the file size.
When you train a machine learning model, the model's weights are quantized to reduce the model's size and improve its performance on embedded devices.
Quantization is a powerful tool that can be used to improve the efficiency and performance of a variety of applications. However, it is important to understand the trade-offs involved in quantization before using it.
Not entirely.Use a calorimeter or photocell to determine E.I'm sure I heard about someone doing that.
It gave the wrong answer.
I think it's because they forgot about chemistry.
Then several labs built water calorimetersWhat was the expected advantage of water calorimeters over the graphite?
The effect of radiation on living tissue is principally due to its interaction with cellular cytoplasm - mostly water. So for radiotherapy, in particular, we would like to know the dose (amount of energy deposited per unit mass) delivered to water by our radiation beam. We can make a theoretical estimate from the electron density of graphite but it would have been nice to eliminate all the approximations inherent in that calculation.Then several labs built water calorimetersWhat was the expected advantage of water calorimeters over the graphite?
Your pixels seem to be binary, whereas real camera pixels are linear.I think you missed to see my previous post.
Image size and intensity also affect the recorded image on the sensor.
(https://www.thenakedscientists.com/forum/index.php?action=dlattach;topic=81120.0;attach=34151)
Here it is. A closer look at sparkling water surface.What needs explaining?
Everything you see is entirely plausible. Not every explanatory hypothesis is plausible. That's science!Do you have an explanatory hypothesis for the observation?
What needs explaining?Anything that's seem unusual from a typical image reflected by water surface.
Anything that's seem unusual from a typical image reflected by water surface.Nothing is unusual, so no explanation is necessary. Easy-peasy.
Do you have an explanatory hypothesis for the observation?Water reflect light.
Anything that's seem unusualWhat do you think is unusual?
Do you have an explanatory hypothesis for the observation?Water reflect light.
The surface of water is disturbed by things like the wind and fountains.
What else needs explaining?
I've identified 6 effects related to saturation and pixelation/spatial quantization of the photosensors. What do you think would happen if those effects don't actually occur?Anything that's seem unusualWhat do you think is unusual?
Everyone else here seem to think that what you see is exactly what we would expect to see.
Quantisation in analogue to digital converters and quantisation of photon energy are entirely separate phenomena.What makes them different?
I've identified 6 effects related to saturation and pixelation/spatial quantization of the photosensors.Well, I'm not convinced about the nomenclature, but the effects seem reasonable.
What do you think would happen if those effects don't actually occur?Since they do, in fact, occur, why should I concern myself ?
I've identified 6 effects related to saturation and pixelation/spatial quantization of the photosensors.Well, I'm not convinced about the nomenclature, but the effects seem reasonable.
So what?
They are not "unusual".What do you think would happen if those effects don't actually occur?Since they do, in fact, occur, why should I concern myself ?
It's like asking what would happen if it rained cows.
In this video Paul Andersen shows you how to construct explanations with evidence in a mini-lesson on Observational Explanations. Two examples are included in the video and two additional examples are included in the linked thinking slides.
TERMS
Explanation - a logical reason for a phenomenon
Observations - a statement about something you notice
Phenomena - observable events in the natural world (require explanations)
Question - a sentence that asks for information
Reasoning - the action of thinking about something in a logical way
This progression is based on the Science and Engineering Practices elements from the NRC document A Framework for K-12 Science Education. ?Make observations (firsthand or from media) to construct an evidence-based account for natural phenomena.?
Source: https://www.nextgenscience.org/
The word has two different meanings (a) representing a continuous function with a discontinuous function and (b) modelling the behavior of electromagnetic radiation as a stream of particles each having a discrete quantity of energy.Quantisation in analogue to digital converters and quantisation of photon energy are entirely separate phenomena.What makes them different?
which doesn't seem to be fully explained by simple reflection from a randomly rippling reflective surface.And again....
They won't be explained if the pixelation and saturation of the photosensors were excluded.which doesn't seem to be fully explained by simple reflection from a randomly rippling reflective surface.And again....
what bits do you think are not explained?
Do you find other effects not yet in my list?
This one shows Sun Reflection on Pacific Ocean from Space.
They won't be explained if the pixelation and saturation of the photosensors were excluded.
Because someone said they are not necessary.They won't be explained if the pixelation and saturation of the photosensors were excluded.
Why would you exclude pixelation and saturation, knowing that doing so would give you the wrong answer
But these are not enough to explain the sparkling effect.I think it perfectly explains the effect.
Do you now realize that the sparkling is not quantized?
Anything that's seem unusual from a typical image reflected by water surface.Nothing is unusual, so no explanation is necessary. Easy-peasy.
Do you have an explanatory hypothesis for the observation?Water reflect light.
The surface of water is disturbed by things like the wind and fountains.
What else needs explaining?
Just to add mischief to your confusion, remember that the displayed image is not usually a 1:1 mapping of the receptor pixel signals. Right now I am using a 3840 x 2160 display - about 8 Mpx - but my camera has a 16 Mpx receptor and the webcam on the computer is only 500k. There are all sorts of software interventions that "hide the joins" and either smooth out or enhance contrast in adjacent areas. In radiology we tend to capture images with the highest available spatial resolution but often "soften" the display so that tiny "punctate" features are more conspicuous - there is an optimum contrast/detail balance somewhere between the maxima of either variable.I'm aware of your concerns here.
That's why I posted many videos from other Youtubers to show variations of camera resolution, their contrast and saturation levels, and resolution of the resulting videos.The presence of other effects, like diffraction, flaring, scattering, and various sizes of the ripple makes it hard to exclusively observe the sparkling effect.In modern digital cameras, including those in smart phones, software filters and image processing algorithms can also affect the results of sparkling effect.
The resolution of the video files are typically lower than the resolution of main camera sensor. In my video, the resolution is 0.92 megapixels (1280 x 720), although the main camera itself has resolution of 50 megapixels. Some image processing must be involved in reducing the resolution.
This movie has been captured with an intensified CCD camera. The movie consists of 200 frames, with exposure times ranging between 0,025 milliseconds and 6,000 milliseconds. It shows how individual photons, transmitted through a double slit, form an interference pattern. It shows wave-particle duality of light.Does anyone notice that the bright spots have various brightness? How should we interpret it?
That's why I posted many videos from other Youtubers to show variations of camera resolution, their contrast and saturation levels, and resolution of the resulting videos.None of which relates quantisation of photon energy to the appearance of sunlight reflected from rippling water.
The relationship lies in how the photosensitive sensors work.That's why I posted many videos from other Youtubers to show variations of camera resolution, their contrast and saturation levels, and resolution of the resulting videos.None of which relates quantisation of photon energy to the appearance of sunlight reflected from rippling water.
Does anyone notice that the bright spots have various brightness? How should we interpret it?The signal delivered by a single pixel irradiated by a single photon depends on the photon energy (which we can assume constant in this instance as a fairly good interference pattern appears) and the time between impact and readout (which is random).
CHAPTERS
0:00 The original paper implied retrocausality
1:23 Really cool metal posters: Displates!
2:37 A classical interpretation would show retrocausality
3:49 How the double slit experiment works
6:25 Debunking the clean double line pattern
7:49 The Delayed Choice Quantum Eraser set up explained
11:54 How the Scientis hand-selected the outcome of the Delayed Choice experiment
SUMMARY
The original paper by the authors who first performed the Delayed Choice Quantum Eraser implied retro causality. But retro causality is true only if you assume a classical way of thinking. But that's not the way quantum mechanics works, and I was wrong for interpreting it that way in my original 2019 video. When viewed with the standard interpretation of quantum mechanics where a particle is always a wave until the moment it is measured, there is no retro causality.
How the double slit experiment works: If you send photons one at a time through the slits, at first you will see what looks like a random distribution of dots. But after a while, you will see that those dots create an interference pattern.
If you then put detectors on the slits to measure which slit the photon passes through, you see a pattern like you would if you were sending individual particles through the slits. The act of measuring seems to affect the results. But the change is due to the nature of quantum mechanics. All quantum objects like photons and electrons are really waves. But if they interact with anything, that is, if an irreversible energy exchange takes place, their waves become localized like a particle. This is called ?wave collapse.? Wave collapse also occurs when the photon interacts with the screen in the back. And we this as a dot on the screen.
The Delayed Choice Quantum Eraser is like the double slit experiment on steroids. First, I want to point out that if you have a detector that measures the path, you don?t really get two clean lines of photons like it's usually illustrated. You get a single spread out distribution of photons.
How does the delayed choice experiment work?
It starts with the double slit, but first the photons go through a special optical device called a Barium Borate crystal. It splits a single photon into a pair of entangled photons with half the energy each of the original. Note that the process of creating entangled photons effectively results in a measurement. In other words, the wave function of the photon collapses so that it is now a particle. And since the path from the top slit to detector 1 is slightly different than the path from detector 2, the which way information of the photon is known. Thus the pattern that will show up at detector 1 will always be a spread out pattern, not an interference pattern. It doesn?t matter what happens at any of the other detectors.
So why is it illustrated as changing depending on what happens at the other detectors? This is the center of the confusion, and where the idea of retro causality comes in.
Well the confusion is from the way this experiment is presented - as D1 changing its pattern to match the interference pattern at D4 or D5 when the photons end up there, but showing a different pattern, a spread out pattern, if the photons end up at D2 or D3.
So this implies that what happens at D2, D3, D4 or D5 influences what happens at D1. But since the path to D1 is shorter than the path to any of the other detectors, the photons reach D1 BEFORE they reach D2, D3, D4 or D5. So the implication is that the pattern at D1 which would be in the past, is being affected by what happens in the future at D2, D3, D4 or D5. So people have naturally been led to think that this means retro causality. This is wrong.
The quantum eraser has no effect on the original screen. What?s really happening is that the changing patterns are due to the scientists, conducting this experiment, selecting subsets of the photons in D1 to show the same patterns as at each of the other detectors. This can be done because the particles hitting the screen at D1 and the particles going to the other detectors are entangled.
#delayedchoicequantumeraser
#quantumphysics
So in the presentations that you see, including the one I originally made, the interference pattern you see get at D1 is nothing but a hand-selected subset of the actual original spread out pattern at D1, corresponding to photons that ended up at D4 or D5. This is done post-experiment by hand! The patterns do not change on their own. The future does not affect the past.
For comparison, double slit experiment using electrons shows more uniform intensity in each bright spot, at least until one spot is hit more than once.Because the electron charge is a constant, unlike the signal you get from an intensified photon detector.
Interference pattern built up photon by photon
Quote
This movie has been captured with an intensified CCD camera. The movie consists of 200 frames, with exposure times ranging between 0,025 milliseconds and 6,000 milliseconds. It shows how individual photons, transmitted through a double slit, form an interference pattern. It shows wave-particle duality of light.
Does anyone notice that the bright spots have various brightness? How should we interpret it?
Moreover, what is the size of the photons producing that bright spots?
Do they depend on their frequency?
Do they depend on their polarization?
Boy, Was I Wrong! How the Delayed Choice Quantum Eraser Really worksIt shows The Problem With Science Communication.
Moreover, what is the size of the photons producing that bright spots?
Do they depend on their frequency?
Do they depend on their polarization?
What determined the diameter of those bright spots on the screen in those single photon double slit experiments?Moreover, what is the size of the photons producing that bright spots?
Do they depend on their frequency?
Do they depend on their polarization?
Not sure what you mean by "size" of a photon. SInce there is a clear interference pattern. the photons will all have had pretty much the same frequency/wavelength/energy.
What determined the diameter of those bright spots on the screen in those single photon double slit experiments?See reply #257 above.
If the frequency of the light source is doubled, would it change the diameter of those bright spots?Generally, yes, because each interaction with the primary detector will release twice as much energy, either in the form of visible photons or electrons depending on the incident radiation and the type of detector used. Most likely electrons.
The pictures I posted here are screenshots of my next videos. The first will show a closer look to the phenomenon, while the second one tries to offer some explanations.Here's the video trying to explain the effects found in sparkling water surface videos.
The existence of several effects at once in a single piece of evidence may prevent us from identifying the phenomenon in the first place.
I'd like to have some constructive feedback to improve my videos before uploading them. Does anyone notice some unusual effects I haven't mentioned yet? Or think that the effects I did mention here implausible?
What determined the diameter of those bright spots on the screen in those single photon double slit experiments?See reply #257 above.If the frequency of the light source is doubled, would it change the diameter of those bright spots?Generally, yes, because each interaction with the primary detector will release twice as much energy, either in the form of visible photons or electrons depending on the incident radiation and the type of detector used. Most likely electrons.
https://www.tedpella.com/cameras_html/ccd_cmos.aspx
(https://www.tedpella.com/cameras_html/ccd.jpg)
Figure 1: Diagram of a CCD.
On a CCD, most functions take place on the camera's printed circuit board. If the application's demands change, a designer can change the electronics without redesigning the imager.
(https://www.tedpella.com/cameras_html/cmos.jpg)
Figure 2: Diagram of a CMOS.
A CMOS imager converts charge to voltage at the pixel, and most functions are integrated into the chip. This makes imager functions less flexible but, for applications in rugged environments, a CMOS camera can be more reliable.
This difference in readout techniques has significant implications for sensor capabilities and limitations. Eight attributes characterize image sensor performance.
Responsivity, the amount of signal the sensor delivers per unit of input optical energy. CMOS imagers are marginally superior to CCDs.
Dynamic range, the ratio of a pixel's saturation level to its signal threshold. CCDs have the advantage here.
Uniformity, is the consistency of response for different pixels under identical illumination conditions. CMOS imagers were traditionally much worse than CCDs, however new amplifiers have made the illuminated uniformity of some CMOS imagers close to that of CCDs.
Shuttering, the ability to start and stop exposure arbitrarly, is superior in CCD devices. CMOS devices require extra transistors or nonuniform shuttering, sometimes called rolling shuttering to achieve the same results.
Speed, an area in which CMOS arguably has the advantage over CCDs because all of the camera functions can be placed on the image sensor.
Windowing, CMOS technology has the ability to read out a portion of the image sensor allowing elevated frame rates for small regions of interest. CCDs generally have limited abilities in windowing.
Antiblooming, is the ability to gracefully drain localized overexposure without compromising the rest of the image in the sensor. CMOS generally has natural blooming immunity. CCDs require specific engineering to achieve this capability.
Biasing and clocking. CMOS imagers have a clear advantage in the area, operating on a single bias voltage and clock level.
CCD and CMOS imagers were both invented in the late 1960's. CCD became dominant in the market, primarily because they produced superior images with the fabrication technology available. CMOS image sensors required more uniformity and smaller features than silicon wafer foundries could deliver at the time. Not until the 1990's, with the development of lithography was there a renewed interest in CMOS. That interest is due to lower power consumption, camera-on-a-chip integration, and lowered fabrication costs. Both CCD and CMOS imagers offer excellent imaging performance. CMOS imagers offer more integration (more functions on the chip), lower power dissipation (at the chip level), and the possibility of smaller system size.
Today there is no clear line dividing the types of applications each can serve. CCD and CMOS technologies are used interchangeably. CMOS designers have devoted intense effort to achieving high image quality, while CCD designers have lowered their power requirements and their pixel sizes. As a result, you can find CMOS sensors in high-performance professional and industrial cameras and CCDs in low cost low power cell phone cameras. For the moment, CCDs and CMOS remain complementary technologies-one can do things uniquely the other cannot. Over time this distinction will soften, with more CMOS imagers consuming more and more of the CCD's traditional applications. Considering the relative strength and opportunities of CCD and CMOS imagers, the choice continues to depend on the application and the vendor more than the technology.
What determined the diameter of those bright spots on the screen in those single photon double slit experiments?See reply #257 above.If the frequency of the light source is doubled, would it change the diameter of those bright spots?Generally, yes, because each interaction with the primary detector will release twice as much energy, either in the form of visible photons or electrons depending on the incident radiation and the type of detector used. Most likely electrons.
https://www.tedpella.com/cameras_html/ccd_cmos.aspx
(https://www.tedpella.com/cameras_html/ccd.jpg)
Figure 1: Diagram of a CCD.
On a CCD, most functions take place on the camera's printed circuit board. If the application's demands change, a designer can change the electronics without redesigning the imager.
(https://www.tedpella.com/cameras_html/cmos.jpg)
Figure 2: Diagram of a CMOS.
A CMOS imager converts charge to voltage at the pixel, and most functions are integrated into the chip. This makes imager functions less flexible but, for applications in rugged environments, a CMOS camera can be more reliable.
This difference in readout techniques has significant implications for sensor capabilities and limitations. Eight attributes characterize image sensor performance.
Responsivity, the amount of signal the sensor delivers per unit of input optical energy. CMOS imagers are marginally superior to CCDs.
Dynamic range, the ratio of a pixel's saturation level to its signal threshold. CCDs have the advantage here.
Uniformity, is the consistency of response for different pixels under identical illumination conditions. CMOS imagers were traditionally much worse than CCDs, however new amplifiers have made the illuminated uniformity of some CMOS imagers close to that of CCDs.
Shuttering, the ability to start and stop exposure arbitrarly, is superior in CCD devices. CMOS devices require extra transistors or nonuniform shuttering, sometimes called rolling shuttering to achieve the same results.
Speed, an area in which CMOS arguably has the advantage over CCDs because all of the camera functions can be placed on the image sensor.
Windowing, CMOS technology has the ability to read out a portion of the image sensor allowing elevated frame rates for small regions of interest. CCDs generally have limited abilities in windowing.
Antiblooming, is the ability to gracefully drain localized overexposure without compromising the rest of the image in the sensor. CMOS generally has natural blooming immunity. CCDs require specific engineering to achieve this capability.
Biasing and clocking. CMOS imagers have a clear advantage in the area, operating on a single bias voltage and clock level.
CCD and CMOS imagers were both invented in the late 1960's. CCD became dominant in the market, primarily because they produced superior images with the fabrication technology available. CMOS image sensors required more uniformity and smaller features than silicon wafer foundries could deliver at the time. Not until the 1990's, with the development of lithography was there a renewed interest in CMOS. That interest is due to lower power consumption, camera-on-a-chip integration, and lowered fabrication costs. Both CCD and CMOS imagers offer excellent imaging performance. CMOS imagers offer more integration (more functions on the chip), lower power dissipation (at the chip level), and the possibility of smaller system size.
Today there is no clear line dividing the types of applications each can serve. CCD and CMOS technologies are used interchangeably. CMOS designers have devoted intense effort to achieving high image quality, while CCD designers have lowered their power requirements and their pixel sizes. As a result, you can find CMOS sensors in high-performance professional and industrial cameras and CCDs in low cost low power cell phone cameras. For the moment, CCDs and CMOS remain complementary technologies-one can do things uniquely the other cannot. Over time this distinction will soften, with more CMOS imagers consuming more and more of the CCD's traditional applications. Considering the relative strength and opportunities of CCD and CMOS imagers, the choice continues to depend on the application and the vendor more than the technology.
How do you think the double in size of the bright spots come out?Scatter within the CCD, scatter within the intensifier, and halo from the optics, are the usual causes.
I was asking about doubling in size of the bright spot shown in resulting image when the frequency of light source is doubled, without changing anything else.How do you think the double in size of the bright spots come out?Scatter within the CCD, scatter within the intensifier, and halo from the optics, are the usual causes.
See reply #264, second part.Do you realize that pixel size is fixed?
Yes, and the bright spots are larger than the pixels.Activating two pixels at once will make the bright spot elongated.
Activating two pixels at once will make the bright spot elongated.The spots you can see on the image almost certainly span more than two pixels.
What would happen if the photon frequency is only 50% higher than before?You get 50% more energy deposited in the first interaction.
The spots you can see on the image almost certainly span more than two pixels.What happens in case of less than two pixels?
You get 50% more energy deposited in the first interaction.In what form?
In quantum mechanics, wave function collapse occurs when a wave function?initially in a superposition of several eigenstates?reduces to a single eigenstate due to interaction with the external world.In various experiments using dim light source, the wave function doesn't seem to collapse when it interacts with many kinds of objects such as mirrors, lens, polarizers, slits, gratings, quarter wave plates, half wave plates, air molecules, and beam splitters. Only certain kinds of objects can cause the wave function to collapse, such as electronic photosensors. There must be something that causes that difference in behavior.
https://en.m.wikipedia.org/wiki/Wave_function_collapse
You get a very tiny dot in the displayThe spots you can see on the image almost certainly span more than two pixels.What happens in case of less than two pixels?
In whatever form the sensor converts it to. Could be ion formation, photoelectron emission, visible photons, or electrons promoted to a higher trap level in a phosphor.You get 50% more energy deposited in the first interaction.In what form?
You get a very tiny dot in the displayOne pixel is the minimum non-zero result.
QuoteIn quantum mechanics, wave function collapse occurs when a wave function?initially in a superposition of several eigenstates?reduces to a single eigenstate due to interaction with the external world.In various experiments using dim light source, the wave function doesn't seem to collapse when it interacts with many kinds of objects such as mirrors, lens, polarizers, slits, gratings, quarter wave plates, half wave plates, air molecules, and beam splitters. Only certain kinds of objects can cause the wave function to collapse, such as electronic photosensors. There must be something that causes that difference in behavior.
https://en.m.wikipedia.org/wiki/Wave_function_collapse
History and context
The concept of wavefunction collapse was introduced by Werner Heisenberg in his 1927 paper on the uncertainty principle, "?ber den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik", and incorporated into the mathematical formulation of quantum mechanics by John von Neumann, in his 1932 treatise Mathematische Grundlagen der Quantenmechanik.[10] Heisenberg did not try to specify exactly what the collapse of the wavefunction meant. However, he emphasized that it should not be understood as a physical process.[11] Niels Bohr also repeatedly cautioned that we must give up a "pictorial representation", and perhaps also interpreted collapse as a formal, not physical, process.[12]
Consistent with Heisenberg, von Neumann postulated that there were two processes of wave function change:
The probabilistic, non-unitary, non-local, discontinuous change brought about by observation and measurement, as outlined above.
The deterministic, unitary, continuous time evolution of an isolated system that obeys the Schr?dinger equation (or a relativistic equivalent, i.e. the Dirac equation).
In general, quantum systems exist in superpositions of those basis states that most closely correspond to classical descriptions, and, in the absence of measurement, evolve according to the Schr?dinger equation. However, when a measurement is made, the wave function collapses?from an observer's perspective?to just one of the basis states, and the property being measured uniquely acquires the eigenvalue of that particular state,
λ. After the collapse, the system again evolves according to the Schr?dinger equation.
Confining the light causes it to have more colours. This is explained well in an excellent video by Ben Miles. But at first glance, the experiment seems to be a totally different one than what I explained in the video- so I'll explain what the connection is. In particular, it's only the "single slit" version of the experiment that's relevant for us. I said that the researchers confined a laser to a small space. The technique they used to do this was to have two lasers- one which is the source, and the other which is used to turn on and off a "switch" of sorts. What the switch does is it makes the material in the experiment go from transparent to reflective very quickly, then back. The source laser is shining continuously at the material.
But the idea is that for the short while that the material is reflective a little section of the laser beam is reflected. That's the "confined" light- they took a laser beam that's always on and constant and isolated a small section, confining the whereabouts of the light. They then measured the colours of that light and find it's spread out. (This result is at the 8 minute mark)
Clearly written by someone with no understanding of physics.Perhaps you just understand it differently than she does. At least her PhD advisors and reviewers agreed that she understand some of it.
Let's analyze the most simple case, and see if our models can make sense of it. An electron is made to oscillate at frequency of 1 Hz. The amplitude is 1 meter. How many photons is it radiating every second in average?IMO, the last question would be constrained by speed of light. But the answer to the other questions are less obvious.
What if the amplitude is then reduced to 1 mm?
Is there a minimum amplitude?
Is there a maximum amplitude?
It's often said that light is an electromagnetic wave, a disturbance in electric and magnetic fields, but what does that mean? How are they made? Let's take a deeper look at electrodynamics and this history behind the discovery to see if we can find an answer.
Is there a maximum amplitude?Obviously.
Is there a maximum amplitude?Obviously.
Since the electron can't move faster than c, then the maximum distance it can travel in 1 second is around 300 million meter, thus the maximum amplitude is 75 million meter.Let's analyze the most simple case, and see if our models can make sense of it. An electron is made to oscillate at frequency of 1 Hz. The amplitude is 1 meter. How many photons is it radiating every second in average?IMO, the last question would be constrained by speed of light. But the answer to the other questions are less obvious.
What if the amplitude is then reduced to 1 mm?
Is there a minimum amplitude?
Is there a maximum amplitude?
It seems like the currently most widely accepted model isn't adequate to give us the satisfactory answers.
You may be about to confuse Planck's "particle in a box" model with a Maxwellian "resonant aerial" model. Beware!I don't assume those models. My assumptions are as follow:
Which model do you think is best to describe my example?Since your example is of an oscillating current at 1 Hz, Maxwell will do the job very well. It is indeed used by those who design VLF antennae for undersea communication.
Story of how Planck discovered the blackbody radiation formula and why he introduced energy quantization as a math trick
Errata:
08:10 instead of Pringscheim should be Pringsheim, thanks to @petermarksteiner7754 for notifying this
14:40 after the integration there is an extra minus sign that should not be there, thanks @escandestone6001 for notifying this
20:22 second equation should be ε/(kT)=log(1+ε/U), thanks to @Galileosays for notifying this
23:52 "gasses" should be "gases," thanks to @Robert-skibelo for notifying this
The cases are distinct and unrelated.Just multiply the frequency by 1 quadrillion times, and you get ultraviolet light. That's a relationship, no matter how hard you try to deny it.
which is mostly dismissed in textbooks.
Macroscopically, the ideal gas law states that, for an ideal gas, the product of pressure p and volume V is proportional to the product of amount of substance n and absolute temperature T:
pV=nRT,
where R is the molar gas constant (8.31446261815324 J⋅K−1⋅mol−1).[3] Introducing the Boltzmann constant as the gas constant per molecule[4] k = R/NA transforms the ideal gas law into an alternative form:
pV=NkT,
where N is the number of molecules of gas.
kB derives from classical statistical mechanics.Do you meant it's no longer needed?
Of course it's needed. How else can you describe the Boltzmann distribution, rationalise the gas laws, or do thermodynamics?How does Boltzmann's constant relate to black body radiation?
Quantum mechanics doesn't displace classical mechanics, any more than relativity displaces newtonian physics. The test of both is that they degenerate to the classical formula for large or slow systems.
Planck's formula and Wien's approximation both attempt to describe the spectrum of black body radiation, the characteristic energy distribution emitted by an ideal perfect absorber at a given temperature. However, they differ in their accuracy and underlying assumptions.
...
In conclusion, Planck's formula is the more general and accurate description of black body radiation, thanks to its inclusion of energy quantization. Wien's approximation provides a useful simplification for shorter wavelengths but is not sufficient for the entire spectrum.
Planck's Formula B(ν, T) = (2 * h * ν^3) / (c^2 * (e^(h * ν / (k_B * T)) - 1))
Wien's Approximation B(ν, T) = (2 * h * ν^3) / (c^2 * e^(h * ν / (k_B * T)))
You're right, at first glance, the equations might seem very similar, with Planck's formula just adding a -1 term to Wien's approximation. However, that seemingly small addition has a profound impact:
The -1 term accounts for quantization: This term incorporates the revolutionary idea that energy can only be emitted or absorbed in discrete packets. This concept of quantized energy levels was a major breakthrough in physics and went beyond the classical framework used in Wien's approximation.
Just shows why you shouldn't use ChatGPT as a source of scientific information. It displays all the insight and understanding of a politician.It reflects the quality of training data sources. It seems like it hasn't been equipped with the tools to filter out incorrect statements.
When the AI models still depend on their human supervisors to select which data sources are more accurate and reliable than the others, then the quality of those supervisors will determine the quality of responses that they'll produce. Leaving their current mistakes uncorrected will make it harder and more expensive to correct them in the future.Just shows why you shouldn't use ChatGPT as a source of scientific information. It displays all the insight and understanding of a politician.It reflects the quality of training data sources. It seems like it hasn't been equipped with the tools to filter out incorrect statements.
anything from an AI source may be infected by untruthHuman sources are not immune from falsehood either. The probabilities for their occurrence can only be reduced by more check and recheck, compared to observations.
But humans tend to conduct experiments or simply ask "does that seem reasonable?" AI just goes on regurgitating any old rubbish because it doesn't care about the consequences of its output.But the developers do care about the accuracy of their AI models. It determines whether or not they can get their return of investment.
The model may be 100% accurate and quote Donald Trump's authoritative speech to the letter, but I wouldn't want anyone to inject themselves with hydrochloroquinone.New AI models will be able to filter out misinformation, based on previously accumulated information, and comparison with various other information sources, including their own sensors.
If all the previously accumulated information is the Bible, AI will reject any scientific statement as unreliable and the world will regress to the intellectual level of the Republican Party.That's a big if. And fortunately it's not true.
It was once. Similar problems have been found where all the permitted truth was all in Mao's Little Red Book, or Lysenko's state-approved botany papers.It would be harder to censor worldwide openly accessible information sources. Especially with VPN and satellite connection.
Problem with the internet as a source is that there is only one right answer to scientific question, but an infinite number of wrong ones, and the internet grants allows them equal weight.
Traditional search engines like Google prioritize most referred sources....like The Bible, Koran, Little Red Book, Shakespeare, Alice in Wonderland, and other frequently-cited sources of utter nonsense.
At best, consensus represents the opinion of the average man.That's what democracies depend on. Democratic societies whose constituents are mostly incompetent are less likely to survive. Usually, least competent members of the societies would die out earlier. It would tip the average value upward.
Advances in understanding are made by exceptional people questioning the consensus.There are some types of research gaps.
In this video, I dive deep into the landscape of academic research. I share insights from my own journey, focusing on the discovery and exploration of research gaps. These gaps, often seen as voids in our understanding, are in fact golden opportunities for innovation and progress. I discuss four distinct types of research gaps that I've encountered, including the personal challenges and triumphs I've experienced in bridging them.
At best, consensus represents the opinion of the average man.Or average researchers.
Embark on a journey to redefine your academic search with Consensus AI, the cutting-edge tool transforming the landscape of research.
TIMESTAMPS
00:00 How to use Consensus
00:48 How to Use AI Summary
02:43 How to use Consensus Meter
03:23 How to use Consensus Co Pilot
04:47 How to Filter with Consensus
06:40 Digging Deeper into Consensus
08:41 Consensus on the GPT Store
09:38 Wrapping Up
Democratic societies whose constituents are mostly incompetent are less likely to survive.It is a characteristic of most human societies that we protect and nurture the incompetent and superstitious, and even insist that they have a right to breed and/or disseminate their superstitions. Hence
Usually, least competent members of the societies would die out earlier.is no longer a "given"
It would tip the average value upward.is clearly not the case in, for example, the USA or anywhere that religion or party/tribal loyalty influences the democratic process.
is clearly not the case in, for example, the USA or anywhere that religion or party/tribal loyalty influences the democratic process.There might be some exceptions in short terms, for whatever reason. But in the long term the trend is usually maintained.
In atoms, only certain orbitals are permitted (quantization)
- To calculate these orbitals, you need to solve the wave equation for the electron.
- Some of these orbitals are spherical, but others look like a cluster of balloons assembled by a clown. How do you calculate the radius and centripetal motion for these?
We describe the possible fundamental vibrations on a sphere in three dimensions by counting, mirroring and rotating nodal lines.
This video ist part of the online course www.quantumreflections.net dealing with quantum physics, produced by the institute for physics education research, M?nster university (Germany)
Spatially, the patterns extend to infinity, although with quickly dropped magnitude for further distances. While dark regions within finite distances might be interpreted as products of destructive interference.Based on the radiation patterns that we can measure for various shapes of antenna where we can control the motion of the electrons, we can calculate backwards to infer how electrons should move around the nucleus to produce radiation patterns resembling known atomic orbitals. It will require adequately accurate and precise electrodynamic model.
Unlike in a dipole antenna where the movement of electrons is confined in one dimension, the electrons around atomic nucleus can move in three dimensions.
Based on the radiation patterns that we can measure for various shapes of antenna where we can control the motion of the electrons, we can calculate backwards to infer how electrons should move around the nucleus to produce radiation patterns resembling known atomic orbitals. It will require adequately accurate and precise electrodynamic model.It is precisely the absence of electromagnetic emission from atoms in equilibrium that put an end to the Bohr atom.
It's as if electrons can magically jump from one region with high probability to another regions without having to travel the space between them.There's no implication that they magically jump. If you start with a classical model, you probably won't derive a quantum result, but if you start with a quantum and probabilistic model, you can usefully predict molecular orbitals that correspond with ball-and-stick models, crystallographic measurements, and observed spectra.
It is precisely the absence of electromagnetic emission from atoms in equilibrium that put an end to the Bohr atom.No. It's why Bohr proposed his model, against Rutherford's planetary model.
There's no implication that they magically jump.You can say that they casually leap or tunnel. But what matters is what it means. It suggests that electrons can disappear from their previous positions and reappear in new position without traveling through the space in between.
You can say that they casually leap or tunnel.No. Tunneling is an entirely different phenomenon.
It suggests that electrons can disappear from their previous positions and reappear in new position without traveling through the space in between.
No. It's why Bohr proposed his model, against Rutherford's planetary model.Bohr's electrons orbit, That's the problem.
He postulated that some orbits are stable, hence electrons occupying them don't radiate energy.No. It's why Bohr proposed his model, against Rutherford's planetary model.Bohr's electrons orbit, That's the problem.
No. Probability does not imply that anything moves.How can electrons have momentum if they don't move?
No. Tunneling is an entirely different phenomenon.Are you fine with the word leap?
He postulated that some orbits are stable, hence electrons occupying them don't radiate energy.Orbiting equals acceleration. Accelerating charges emit electromagnetic radiation. So the classical model is wrong.
Different phenomena.No. Tunneling is an entirely different phenomenon.Are you fine with the word leap?
Which phenomena are you referring to?Different phenomena.No. Tunneling is an entirely different phenomenon.Are you fine with the word leap?
Accelerating charges emit electromagnetic radiation.Not necessarily. Circulating current in a ring superconductor can stay for years with no perceivable decay.
Bohr's model is not usually considered classical.He postulated that some orbits are stable, hence electrons occupying them don't radiate energy.Orbiting equals acceleration. Accelerating charges emit electromagnetic radiation. So the classical model is wrong.
Which phenomena are you referring to?Jumping and tunneling.
Which is why you need quantum mechanics to explain superconductivity.Accelerating charges emit electromagnetic radiation.Not necessarily. Circulating current in a ring superconductor can stay for years with no perceivable decay.