If light contains particles called photons, perhaps they should have momentum like any other particle...here momentum = E/c=hf/c
i think frequency of photon does not change in calculating momentum of a photon...therefore knowing position is enough and momentum is full of constants...so i can say Heisenberg's uncertainty principle is not a big issue to study photons .....
Heisenberg has just re-labelled the properties of the Fourier's transform, already known a century earlier.
No corpuscles exist.
Willy Lamb : "There is no object such as "a photon".".
Citation :
Every photon has an absorber.
A photon is a successful transaction between three partners: an emitter, an absorber, and the space or optical devices between them. This transaction transfers by electromagnetic means, a quantum of looping h, and an energy-momentum that depends on the respective frames of the emitter and the absorber.
We distinguish the macro-time of macro-systems such as the laboratory, from the micro-times where dwell all the gropings of broglian waves from which emerge the successful transactions.
Fifteen surreptitious, copenhaguist and corpuscularist postulates we do not more accept.. Available from: https://www.researchgate.net/publication/317605373_Fifteen_surreptitious_copenhaguist_and_corpuscularist_postulates_we_do_not_more_accept [accessed Jun 20, 2017].
Article Fifteen surreptitious, copenhaguist and corpuscularist postu...
@Jacques Lavau
my dear Jacques Lavau i asked question
Do we require Heisenberg's uncertainty principle to study light or photons?
We need real laws of physics.
Maxwell equations for instance, revised for bosons.
The uncertainty principle concept helps with photons.
For a light ray of just one frequency the energy time versión is usefull.
You can get a single frequency if the duration is long, a long light train.
If the duration is short, a very short pulse of light you must need a oombinationof many frequencies. Basic Fourier analysis, closely related to uncertainty principle.
Maxwell equations do not include the bosonic pinch. Do not predict the success of large base interferential astronomy.
My dear Jacques Lavau
Which the emitter emits?
What are electromagnetic means?
@Jose Garrigues : have you ever heard about the equations of Maxwell ? They are of good use in electrotechnics, radioelectricity, radar, hertzian beams, spectrography, optics, radiocrystallography, so on.
If you want photons (bosons) then I suggest you use Quantum
Optics and not a Classical theory such as Maxwell.
When there is only one photon, the bosonic character does not work very much, and for calculating the geometry of the Fermat's spindle between emitter and absorber, knowing the wavelength and the distance, the Maxwell's equations fill the bill.
@Jacques Lavau
Something I have heard of Maxwell's equations.
I am enclosing a prologue written by my colleague Luis
If you want to know what a true photon you can read my book
Classic 4D Model Of the Discrete Universe
https://www.researchgate.net/publication/315837723_Classic_4D_Model_Of_the_Discrete_Universe_C1C2_and_C3
Or you can read
Wave-Particle Duality as a Classic Phenomenon
https://www.researchgate.net/publication/317341594_Wave-Particle_Duality_as_a_Classic_Phenomenon
Book Classic 4D Model Of the Discrete Universe C1,C2 and C3
Article Wave–Particle Duality as a Classic Phenomenon
It is true that uncertainty principle like inequalities can be obtained just
using Classical waves, but you need to know where and how to insert the
Planck constant in order to do this for QM
Additionally the Fourier transform is not the only way to obtain such inequality,
Operator noncommutation is an alternative, perhaps even more commonly used method.
It is well explained in the second volumen of Walter Greiner, quantum theory, or in the well known Landau Lifschitz text.
Walter Greiner, hmm...
Noodle trajectories : http://deonto-http://citoyens.deontolog.org/index.php/topic,887.msg3175.html#msg3175
And page 151 § 10.3 of
http://jacques.lavau.deonto-ethique.eu/Physique/Microphysique_contee_juin2017.pdf
Such blunders because nor Greiner nor Feynman himself took care of h, and its consequence : the intrinsic frequency mc²/h for all "particles" with mass.
http://citoyens.deontolog.org/index.php/topic,887.msg3175.html#msg3175
http://jacques.lavau.deonto-ethique.eu/Physique/Microphysique_contee_juin2017.pdf
It is true, the wave function is in general a complex function.
Most of the time you can get along with just real functions,
However when a magnetic field is introduced you are forced to use a complex wave function. If you are calculating currents, also use complex wave functions.
The wave function is always indeterminate within a phase factor
exp(i alfa) , because with this factor you get the same probability density.
In 1926 Erwin Schrödinger conceived an equation as a way to describe the wave behaviour of particles of matter.
Since then there have been several attempts to obtain the Schrödinger equation of Different principles. If the wave function is related to the probability of finding the particle at a position and time, how is it possible that in almost 100 years after its publication they do not know how to deduce it?
Defining the wave function as “the ratio between the kinetic energy that the electron has when it is unobserved and the energy that it acquires due the observation” it's very easy to deduce.
Derivation of the Schrodinger Equation from Classical Physics
https://www.researchgate.net/publication/303495740_Derivation_of_the_Schrodinger_Equation_from_Classical_Physics
It is clear that the correct interpretation is what most believe
The wave function is real
Article Derivation of the Schrodinger Equation from Classical Physics
You do not have a physical description of the physical device you call "observation". The mystics replaces the physics, and it is so so for 1927.
The observation is made by photons and also I have the physical description of the photon. The current theories are mathematical and non-physical
So don't be surprised when the yield of the scientific teaching si so low. You exempt yourself from any physical axioms, from any semantic axioms, from any experimental disicipline.
I cannot see the benefits of such a complete failure. Sure, there was a territorial benefit for Niels Bohr and for the Göttingen boys as they eliminated two timid and polite competitors, by sheer violence. But now, the beneficiaries are all dead.
Jose
I think if the electron wave behaved as a real classical wave,
a real function as Jacques would maintain, then I think the motion of the electron in this real wave would open up a gap in the free electron spectrum, similarly to a charge density wave. This effect is very familiar in Solid state physics.
For free electrons in a constant potential,
The density would be modulated, instead of being uniformly distributed., that is without any interference effect., or special boundary conditions. Plus it would generate a real detectable frequency.
Since nothing of the kind happens, I just think of a probability density associated with a Complex wave function., and uniform distribution of density for free electrons, unconditioned by a boundary condition.
In certain situations, with boundary conditions, you can have these effects.
This is because probability of finding an electron does have incidence in the physical distribution of these.(ie undulator)
The Dirac-Schrödinger frequency 2mc²/h of the electron has been measured at the ALS : http://aflb.ensmp.fr/AFLB-331/aflb331m625.pdf
Experimental observation compatible with the particle internal clock
http://aflb.ensmp.fr/AFLB-331/aflb331m625.pdf
Your frequency sounds way to high.
If the electron goes click click every time it goes by a wavelength
or half wavelength, it would give a frequency dependent of the
kinetic part alone, as in 2(m -m0)cc/h, consistent with p=h/ lambda
Read the Nobel lecture of P.A.M. Dirac : he mentions that the Zitterbewegung gives the exact Bragg's spatial frequency to account the Compton scattering. I had to rediscover and to re-prove this in 2011, as nobody noticed that.
https://www.researchgate.net/publication/313838399_The_Zitterbewegung_key_of_the_electron-photon_scattering_under_the_laws_of_Bragg_and_Compton
Article The Zitterbewegung : key of the electron-photon scattering u...
Well, Dirac thought that according to his theory the electron speed was c,
which is clearly wrong.
then patch this up zitterbewegung was invented, so at each instant the
electron keeps this trembling motion with c but an average unidirectional speed with the
correct value v
However this frequency would not have to do with the wavelenth, I think
On further análisis, it seems that Dirac incorrectly took a derivative to find his
speed, and the speed is finally found correctly., after squaring the hamiltonian, and then taking the derivative.
Your frequency is consistant with a direct transition beteen dirac positive
and negative energy states.
I do not like th zig zag discussion, for exaample in roger Penrose, road to
reality. Maybe someone can explain.
Either Dirac and Schrödinger, and the team driven by Michel Gouanère, and myself are wrong, either the Göttingen-København pack and their heirs are wrong.
But you have to prove your point.
Jacques
If you read the index of my book you will see that I only use a hypothesis. “Everything (and at all times) verifies Heisenberg's uncertainty principle”
From the particles to the universe passing through the stars, black holes, etc.
The wave function is: (mv2 / 2) / (h / 2t)
m is the mass of the electron calculated from the speed of light
v is the speed of the electron in the hydrogen atom calculated from the speed of light
h is the Planck constant calculated from the speed of light.
t depends on energy
What is the axiom exempted?
If energy is applied to the electron (collision with a photon, magnetic field, etc.) the total energy verifies the principle of Heisenberg uncertainty
If you read the article “wave-particle duality…” you will see that it first calculates the frequency 2mc² / h and from there calculates the charge and the mass of the electron.
The gravitation of Newton is obtained from the principle of uncertainty of Heisenberg. Etc.etc.etc.
@Jose Garrigues. What is the symmetry of the hydrogen atom in its ground state ?
an s wave, spherically symmetric, n=1,l=0
Jose,, cannot understand very much
The teals, Anas crecca, the mallards, Anas platyrhynchos, the northern lapwing (Vanellus vanellus), and a few other birds have interferential colors. The colors change with the incidence. It gives some more minorants for the minimum diameter of a photon. How to make these facts compatible with the corpuscularist mythology ?
The anti-reflect coatings keep some efficiency at medium incidences. Again giving minorants of the diameter of a photon far from its absorber.
So on. Lots of facts.
I suppose n and l are the main and secondary quantum numbers in the hydrogen atom.
The 0 is a mathematical concept, not a physical one. l= 0 you have no physical meaning. The electron in the hydrogen atom is in the state of minimum energy and therefore can not lose energy.
The law of Coulomb Kq2 / r2 = mv2 / r
Kq2 / v = mvr
L = mvr is the angular momentum and Kq2 / v = Le angular electric momentum to distinguish it from L
All the particles verify the principle of uncertainty of Heisenberg (angular momentum)
Bhor quantifies the orbits mvr = nh with which modifies a part of the equation and loses half of the solutions. The two parts of equality must be modified
M v nr = nh but there are two variables r and v therefore M v / l lnr = nh
N = layer number and n = 1 ... N, l = 1 ... N
Layer 1 n = 1 and l = 1 n and l therefore coincide L = Le which gives rise to a circular orbit with two possible electrons (spin, s = + 1/2 or s = -1 / 2)
It is all calculated according to the speed of light
- K constant of the law of Coulomb
m Electron mass
q charge of electron or proton
h constant of Planck
v velocity of the electron the hydrogen atom and coincides with the velocity of the electron in the free state
So the distance r between the electron and proton is determined. Therefore, if the energy of the electron is not modified, it will rotate around the proton.
For a value of r there are infinite combinations x, y and z But the Universe does not affect it, the forces vary with r
And evidently the energy of the electron in the hydrogen atom is calculated.
ewton and Coulomb are only valid at rest (m respect of m ' and q with respect to q', electron and proton). You have to take into account the movement.
Kq2 / r2 (1 + v2 / c2 = mv2 /
This causes that in each return there is a advance as in perihelion of mercury.
This advance causes a non-flat orbit and therefore the electron moves on the spherical surface of radius r and at speed v
Layer 2 N = 2 n = 1,2 and l = 1,2
N = l = 1 spherical orbit 2s2
Three combinations remain, resulting in six electrons 2p6
N = 2, l = 1 and n = 1, l = 2 Elliptic orbits
On the other hand, quantum mechanics does not calculate anything, it measures the energy and solves an equation that it does not know to deduce and affirms that the wave function (the square ) is the probability density
@Jose Garrigues .
With your corpuscular and planetary hypothesis, the electronic capture is absolutely impossible.
However the excited state of 57Fe comes from electronic capture from cobalt 57. Hence the use of 57Fe in Mössbauer effect.
The quantum number l goes from 0 to n-1
l=0 does make sense in QM ! The orbital then has spherical symmetry.
Angular momentum is AN OPERATOR IN QUANTUM MECHANICS
What anyone thinks about a physical object doesn't matter. A photon is a massless particle, that does carry energy and momentum. This means E = |p|c. This is true in free space. When trying to measure its properties one introduces devices that modify it-e.g. in a cavity.
A photon is the quantum of the electromagnetic field-an electromagnetic wave of frequency ω has energy E = N hbar ω and momentum |p| = Nhbar |k| c and describes the state of occupation number N for the photons.
Heisenberg's uncertainty principle refers to the distribution of the value of the momentum about these, average, values. For if one wants to measure the state of the electromagnetic field, one will be measuring the shot noise of the photons-that's how Heisenebrg's uncertainty principle enters the picture in this case, when one attempts to resolve individual photons. And that's been studied for decades.
More generally, the uncertainty principle is relevant whenever one attempts to measure the eigenvalue of the Hamiltonian. Even if the Hamiltonian has a discrete spectrum, like the electromagnetic field in a cavity, any measurement will have an intrinsic error given by the uncertainty principle. This is a standard exercise for any discrete spectrum, in fact; the only relation that changes is the dispersion relation. For the case of a continuum spectrum more care is required, to take into account the normalizability of the states.
Juan
Effectively l = 0 is spherical symmetry in QM
The physical zero is 1-1 = 0, x-y = 0
The Legendre polynomial contains l (l + 1) with l = 0,1,2 ... if we make the change L = l + 1, result (L-1) L with L = 1,2,3 then the spherical symmetry is for L = 1, but since it is a mathematical (non-physical) solution, start in ZERO
Angular momentum is a mathematical operator in QM is not a physical something
L = ih / 2 pi partial derivative of a dimensionless function
It is about finding a mathematical function that matches the measured value
What is the electric charge? What is mass? What is frequency?
The QM starts from the measurement of these properties, while I explain and calculate those properties. The QM is a mathematical theory while my model is a physical model that explains the quantum phenomena and also explains what does not explain the quantum mechanics
Jacques
The QM does not know how to calculate the energy of the electron and also can not deduce the equation of Schrodinger, under these conditions can be realized the electronic capture. If you deduce the Schrodinger equation and calculate the energy of the electron you can not perform the electronic capture. Why?
If the electron is a wave, although the particle is always measured, electronic capture can be performed. If the electron is a particle that generates a wave, electronic capture can not be performed. Why?
The QM does not know how to calculate the properties of the electron, so the electronic capture can be performed. If you explain and calculate the properties of the electron can not perform the electronic capture. Why?
If the constants are due to chance you can perform the electronic capture. If you calculate the constants and the relationships you can not perform the electronic capture. Why?
Which is the reason
Jose
Your model looks something like the Bohr model which preceded QM.
Among other things it did not properly explain l=0 states.(s wave)
You have to answer your own many questions by Reading basic Physics texts
and the history of the quantum. theory.
ie Banesh Hoffman, the strange story of the Quantum, there are many other
popular boks on the subject.
A standard undergraduate text is Arthur Beiser, Introdution to the concepts of
modern physics.
It is a shame people had to resort to so abstract explanation, but that is the way it is.
Well,let me try to answer your question about electron capture ,also a wave (or rather being assiciated with a wave)
On a screen , experiments are done, where electrons impinge as points.
However the experiment is done with the interference of two separate electrom beams
coming from the same source.
You gather data and the points form an interference pattern. So there are both points (capture) and there is also wave behaviour, the interference pattern formed by the points,
Interprate what you will, but that is the gist of the matter.
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