PHOTON INTRINSIC EXPANSION

I used the following Ampere rating formula below for determining the magnetic field strength repulsion rather than the attraction between two wires carrying one Ampere of current for a distance of one meter and spaced one meter apart that generates a force of one Newton.
A Newton is defined as the force required to move one kilogram of mass one meter in one second per second..

Fˇ1 & 2 = 2 kˇm x Iˇ1 & Iˇ2 x delta L ⁄ R

Fˇ1 & 2 = Force in Newtons
2kˇm = Magnetic Constant 10ˆ-7
Iˇ1 & Iˇ2 = Current in wires
Delta L = Length of wires (force fields) in meters
R = Distance between wires in meters.

I could not post the illustration of this Ampere definition.
Most physics books should have this illustration.

Although this formula is used in defining the ampere, it can be used for determining the force between wires or one wire as well.
Now if we consider using this formula for determining the force generated by an electron as it jumps from one orbit to an inner orbit, we can get an idea of the repulsive force within the photon pulse Fˇ1&2. First, we use only one wire (electron path) by dropping the multiplier, the subscripts 1 and 2 and one current Iˇ2. We use the electron coulomb current rating of 1.6x10ˆ-19 and then we determine the distance of the electron path as it moves between orbits by taking its average velocity during a time period of one half wavelength (photon pulse) because the electron moves in one direction only for a single polarity. For the wavelength we use 4.86x10ˆ-7 meters (Hb). Delta L would be 6.65x10ˆ-10 meters for the length of the electron trajectory.

Fˇph = 10ˆ-7 x 1.6x10ˆ-19 x 6.65x10ˆ-10 ⁄ 2.43x10ˆ-7 = 4.38x10ˆ-29 ⁄ 3 N

Fˇph = 1.46 x 10ˆ-29 m/s/s

I divided the force by three (bottom figure) because the average change in velocity is one third of the total average velocity of the electron. The photon pulse of one half wavelength replaces R in the original formula.
If we use that figure to consider the expansion of the photon to be 1.46x10ˆ-29 m/s/s, then we can determine the amount of seconds it would take to double the length of the photon pulse to make Z equal one

Z = λ⁄2 = 2.6 x 10ˆ14 seconds or 2.6 x 10ˆ4 billions of years.

This figure is entirely too low for a reasonable cosmological redshift but remember that this is a force moving a kilogram of mass the distance of the photon pulse. However, it does prove that a photon pulse does have an intrinsic expansion that can move a weight.

The weight does not exist in free space. So the photon pulse is expanding in free space at a much greater rate that would be thousands of times greater to erase the above figure. I will have to come up with a figure for this expansion in free space later.

This mathematical interpretation of the photon intrinsic expansion is additional proof besides the other empirical evidence like the basic electric and magnetic field patterns plus the reality of the existence of the electric motor that uses these force fields should be convincing enough for the skeptics that refuse to accept this photon expansion.

__________________
aka Michael Cyrek

The use of the word "proof" in this context is wholly unacceptable. You have in fact "proven" exactly nothing, except that you can plug some numbers into an equation and do the arithmetic. But one might be excused for asking the obvious question: So what?

Why should we accept that the equation given is a valid model for the physical process? Can you show that, by using this equation, you can reproduce measurements of force applied by photons in radiation pressure? And what is the physical interpretation of the "force" as an electron changes orbits? After all, this is a purely Newtonian concept applied, where Newton's physics does not apply, so it certainly appears to be idle fantasy more than anything remotely "scientific".

Having seen your "proof", I think it's quite likely that the "sceptics" are even more skeptical now than they were before.

cyrek reply

Tim wrote
Why should we accept that the equation given is a valid model for the physical process? Can you show that, by using this equation, you can reproduce measurements of force applied by photons in radiation pressure? And what is the physical interpretation of the "force" as an electron changes orbits? After all, this is a purely Newtonian concept applied, where Newton's physics does not apply, so it certainly appears to be idle fantasy more than anything remotely "scientific".

reply
I said my modification of the 'Ampere rating formula' illustrated in most introductory physics books is the source of my equations.
Are you questioning this defined empirical evidence?

These two fields that would be attracting or opposing each other depending on the direction of current flow are used in the 'repulsive setup'.
The removal of one wire does not make the field around the other wire disappear. You are familiar with the basic nature of these magnetic fields.

The formula ia amended to comply to the real nature of the 'planetary concept' of the HA. I still accept this idea because it is the most probable and realistic.

Electrons in motion generate magnetic fields. You know that.
The electron has a defined distance of travel during these photon emissions. The photon pulses have specific wavelength dimensions.
This is the criteria I use.

The magnetic field patterns, like the electric field, are expanded between the poles and charges. You know this. This expansion is caused by an 'intrinsic force'.
This is the basis of my aregument.

This displaces the EoS concept because there is no 'empirixal evidence for its support.

__________________
aka Michael Cyrek

"Planetary concept" of the Hydrogen atom?
Can you show us what is wrong with the solution provided by Quantum Mechanics?
Considering that this solution is well supported by experimental evidence, what makes the "planetary concept" "most probable and realistic"?

Why do you use these "criteria"?
Why would you use Ampere's formula in the first place?

You already made it clear elsewhere that you do not believe in the expansion of space.
And you already showed a non-scientific attitude towards empirical evidence.
What makes you think that your "argument" has any scientific validity?

__________________
papageno


"Why waste time learning, when ignorance is instantaneous?" - Hobbes (Calvin and Hobbes)

"It's all about context!" - Vince Noir (The Mighty Boosh)

"I've never heard of such a brutal and shocking injustice that I cared so little about!" - Zapp Brannigan (Futurama)

"...because the logic of the lines traced from reality is as poor of aesthetic value as it is strict in consistency. " - Paolo Bozzi (Naive Physics - free translation)

Per this logic, bigfoot also displaces the EoS, doesn't it?

I think there is a giant monkey in 12th dimensional space that is playing with the dough of the universe. This also displaces the EoS, however unlikely. I am RIGHT! 8)

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

Cyrek reply

Papa
The Bohr atom explains the HA fairly well. QM is more applicable to more complex atoms or combinations.

The electron and the proton are particles well tested and explained. You have two particles that attract each other because of their unlike charges. These particles would naturally form a binary just like gravitational bodies. How else could they possibly exist?
This is an absolute certainty. There is no other way they can exist except in isolation as separate particles at higher temperatures .

Can you explain how QM would show the formation of the HA? I do not believe it would be any different.

I use the criteria specified because the Ampere formula determines the force between two currents set at a specified distance and determined for a certain length of the force fields that result from the currents involved. This establishes a relationship between the electron and the forces it produces.
This gives me an opportunity to determine the forces generated by adjusting different values for different conditions as I have specified.
I wanted to find out how strong a magnetic field a single electron would generate. This formula gave me that opportunity.

My argument has scientific validity because I base it on real experimental conditions.
There is NO EXPERIMENTAL evidence that supports the EoS. It is a fabricated concept to replace the Hubble Doppler observations which HAD to be replaced.
So, SHOW ME THE EVIDENCE for the EoS.

Travis
GET REAL.

__________________
aka Michael Cyrek

The Bhor model of the atom explains it quite nicely, until you try to get more precision than the model is capable of producing. It is only an approximation, like QM, to some deeper underlying physics.

Describe to me the force that keeps the negative electron from crashing into the positive proton...

Oh, by the way. Some one who's name escapes me at the moment, had deduced that when GR was applied to the universe as a whole expantion occurs as a result, at approx. the rate which we have 'emperically' deduced. This was before Hubbles observations.

And calm down. Sarcasm is crude, but not as crude as yelling.

I laugh at this because I have taken two years of electronics. You cannot re-vamp those equations to suit a particular need. They are only applicable to the use they were designed for.
Sure, a wrench works like a hammer... But you'll agree (maybe not) that it is not the tool for the job.

And, if you don't mind my saying, you get real. I'll live in a phantasy world/universe when I am good and ready to give up and die.
I can say what I want to, even if I'm not serious. It's America. If you aren't from around here, I FORGIVE YOU, and whatever educational system that spawns such backward thinking.

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

You really have to get away from the Bohr model of the neutral hydrogen atom and use QM. If you don't then you're going to have an interesting time deriving the 21 cm emission band (among many other properties) beloved by astronomers.

(For the more mathematically inclined have a look at
http://tesla.phys.unm.edu/phy537/8/node2.html
to see how it's done, but be warned that it may not be suitable for the faint of heart. )

You also need to be aware that the definition of the Ampere applies to two long straight wires. The field generated by a single point charge will not be the same as that generated by current in a (infinitely) long wire. You might want to try some background reading on the Biot-Savart law (it is from this that you can derive the field due to an infinitely long wire, as assumed in the definition of the Ampere,) as well as, possibly, Maxwell's equations.
[edit]Just a minor additional point. B-S is really a steady state equation, and by itself doesn't incorporate any of the propagation time effects, so even this would only be an approximation (though a far better one.) [/edit]

Cyrek reply

Fortis
The 21 cm wavelength comes primarily from intergalactic deep space where most of the
atomic HA’s are present.
My theory is that a wayward electron passes a HA and causes the proton to flip.
Interstellar and intergalactic space is loaded with free electrons that have been blasted out of the stars during the eruptive explosions of the star flares. This causes the protons magnetic field orientation to flip and this causes the electrons orbital dimension to change, resulting in a gradual adjustment to cause the 21 cm radiation.

I depend more on visualization rather than mathematics. Math is necessary sometimes but a visual understanding is more important because you than understand how things work.

Your second post mentions infinately long wires.
The length of the force measured on these wires as I have explained is one mater in length.


Travis
In a field of ground state HA’s, they are all in the same balanced state between the proton magnetic fields and the electrons magnetic fields to repulse each other to keep the electrons in suspension.
In this state, they are all radiating the same standing wave energy which they do not lose because all the atoms are doing the same thing. Since they are all doing the same thing, there is no interchanging of radiation to cause them to lose energy. In this ground state, the radiations are all sinusoidal and continuous.
Only when electrons are ‘bounced up’ to a higher energy state to absorb a photon, do they then radiate a photon.
Electrons do not exchange energies in the standing wave format.
That is the way I see the stabilization of the nature of the Ha’s.

The formula I used is an application of reducing the Ampere rating formula relationships to a ‘microscopic level’.
The currents are reduced to a single electron. The length of the wires are reduced to the distance of travel of the electron during its transition. The distance between the wires (one meter) is reduced to the wavelength of the photon.
These are mere reductions of the relationships between the electrons magnetic field strengths for a measured distance, their interaction between two fields which I reduced to one field which is still relevant for calculation of this one individual force.

__________________
aka Michael Cyrek

Ok Mike.

__________________

Feynman
>~~~~<
Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool.

Religion is a culture of faith; science is a culture of doubt.

Both visualization and math are important. Sometimes math is necessary to understand too and it is almost always necessary to make predictions, rigorously test, and build a theory.

The funny thing is that QM predicts (correctly) the wavelength of the emission. I don't yet see any quantatitive predictions from your theory. (For example a 21 cm emission from neutral H atoms.)

Visualisation is good to have, (and you can gain physical insight from it), but physics is a mathematically based science. (A physicist and chemist are arguing. "You should use the kilogram as your unit os mass, as this is the defined SI unit," the physicist says. "Not at all," says the chemist, "The gram is a far more useful unit, so this is what you should use." A biologist wandering into the conversation asks, "What are units?" )

You need to:
a) Look at the definition of the Ampere, and
b) Read up on the Bio-Savart equation to see where it comes from (If you want a link to an explanation let me know. )

cyrek reply

Thanks Travis.

dgruss
I do not dispute what you say. A picture is worth a thousand words. Mathematics is important too. But math is limited only to those that understand the language. Everybody can understand a picture.

fortis
Can you draw a mathematical picture of how the atom radiates this wavelength?

My visualization sees an electron circling a proton (planetary style) to cause the proton to spin. This proton spin then generates a magnetic field. The electrons magnetic field resulting from its orbital motion interacts with the protons magnetic field. These fields are repulsive towards each other to stabilize the electron into a ground state orbit. As long as there are no nearby charged particles, this balanced state will remain that way.

There are free electrons in all parts of space that have been blasted out of the stars during the flaring activity. A wayward electron could be passing a HA in space to knock the electron to its deepest lowest orbital energy state which would weaken the electric attractive interaction between the two particles to make the electron radiate a 21 cm photon as it slowly returns to its ground state.

Physics is a product of experimental research that established observational behavior of the physical nature of particles and their forces that cause all the energies. The observations preceded the mathematics. Math was made to fit the observations and this gave it the ability to predict future experimental probabilities.
Granted, observations can be misinterpreted. But when the probability factor is included, the correct results can be accepted
I am inclined to believe that math has lost some of its credibility when abstract theories like ‘string’ and ‘inflation’ theories are accepted.
I use SI units because math uses SI units.
.
I understand that chemistry also uses QM but I am not interested in chemistry.

I am not interested in the details. The electron is the smallest fragment of electric current. It is measured by the strength of its magnetic fields in its electric circuits.
My formulation of the electron at the atomic level is a proper analogy of the ampere definition. I have proven that there is an intrinsic force within the photon fields to justify the intrinsic expansion.
Is the Bio-Savart equation necessary to understand the defined experiment of the Ampere?

__________________
aka Michael Cyrek

Have a look at the link in my earlier quote which contains a derivation of this using QM.

But your visualisation doesn't seem to be able to make any quantatitive prediction of the 21 cm wavelength, whereas (as shown in the earlier link) QM does.

Your model would need to explain why the radiation is at 21 cm and not, for example, 22 cm.
Also, you talk about orbitals. If you look at the orbitals associated with higher angular momentum states (such as d, and, f) you'll see that they are difficult to visualise within the simple planetary model of the H atom.

Normally in terms of a bulk property, i.e. your classical current is a (relatively) uniform density of electrons moving with some velocity. It is not a single point charge.

Do you agree that the definition of the Ampere requires long wires? It is akin to asking what the electrostatic field due to a point charge is, and answering with the field for a long line of charge. (The first has a 1/r^2 behaviour, while the second does not.) It is not the appropriate expression to use to calculate the field due to a moving point charge.
Yes, because this equation allows you to calculate the field due to a current in a long wire. (And also allows you to see how this changes as you make the wire shorter.) Even then, it is only an approximation and to do things properly you should use Maxwell's laws. (Due to the finite speed with which the field propagates. )

Wrong, utterly wrong. First, electrons do not "circle" nuclei in "planetary style". By quantum mechanics all we have is a wavefunction that results in a probability cloud where at best you can calculate the probability of observing an electron with those quantum numbers at that location in the atom.

Also, protons, neutrons, and electrons have an intrinsic spin of 1/2 h-bar, no more, no less. They do not need one another to "cause" one another to spin. There are various spin-orbit and spin-spin interactions that can change the direction of spin, but not the total magnitude.

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

Excuse me if I jump in to ask a question that has tortured me for years--does anyone know what that spin rate actually is, in terms of actual rotation/frequency? I don't get what the whole h-bar thing really means, in explicit physical terms. Thanks!

Classical mechanics works well, except to compute the energies of small sets of particles. On the contrary, the PRINCIPLES of quantum mechanics lead to absurdities (the paradoxes), or numerical errors ( the spontaneous emission is twice larger than the result of quantum mechanics; it is impossible to define a wave function for the photon...).
Quantum mechanics says that it solves the wave particle duality leaving the physicist choose arbitrarily what he likes (or remain in the vague, the users of quantum electrodynamics using optical modes which are not defined).
The solitons solve this problem, giving a mathematical support to the "double solution". of de Broglie.
The FORMALISM of quantum mechanism works well, but it may be considered as a phenomenology, valuable as well in a classical scheme than using the absurd principles of QM.
At the beginning, de Broglie and Schrödinger introduced waves to compute the energies, but it is equivalent and much easier to use Lie algebra; the spin becomes a mathematical concept which is not easily physically explanable.

__________________
JMB

How so?

Some people say that the wave function of the photon is the electromagnetic wave; other explain that it cannot be, but they are unable to propose a true psi wave. What is your choice ?

Mine is that the probability to detect a quantum of energy is proportional to the square of the electric field in the mode that the receiver is able to amplify. As the fields in a mode depend only on a real coefficient (the amplitude), the field necesssarily includes the stochastic field (zero point field at 0K). As the electromagnetic field is linear in the vacuum, it does not have any singular point able to be considered as a particle. (with non-linear equations, there are solitons)

__________________
JMB

I was afraid you'd say that.

But here's where that seems like a dodge to me--we have precisely calculated and utilized the -precession- of the proton magnetic moment, the basis of MRI technology, so why can't we figure out the rate of the inherent spin? If we know the charge on the proton, and the magnitude of its magnetic field, why can't we determine the rate of rotation?

What has been said above sounds like 'we know it's a precise number that doesn't change, but we don't have any idea what it is,' or maybe 'we say 'spin,' but it's not really spin, it's something somehow related but different.' What gives? It seems that if protons are spinning, it would be extremely important to understand how fast.

One way of looking at this is to say that experiments indicated that the electron behaved as if it had an intrinsic angular momentum (and an associated magnetic moment), with a z-component that could be +-hbar/2.

The problem occurs when you try to interpret this in terms of a classical spinning object, such as a sphere. As the electron looks pretty much point-like as far as anyone can determine, this would imply, to use a technical term, a humungous angular velocity. It is simpler to attribute spin angular momentum as being as much of a fundamental property of a particle as mass or charge. This may not be the answer that you were looking for, but hopefully it helps.

cyrek reply

fortis
Regarding orbitals, the planetary model is valid. I have explained this before and I will repeat.

The proton and the electron are the only two particles that can exist in isolation in open space or as an atom.. These particles have been thoroughly investigated and everything is known about them. Throughout space, there are no stationary particles because of the interacting forces. Two particles that are drawn to each other because of their interacting attraction will form a binary such as is seen by the major objects such as stars and the resulting planetary systems.

The proton and the electron will do likewise if their motions slow down enough for this to happen.
Scattered matter will form from the plasmas eventually. These two particles will form a binary. They will radiate a continuous standing wave when in a cold environment. This would be their ground state.
The electron, in open space, will spiral into the proton if in a higher energy state until its velocity and the protons subsequent spin resulting from the electrons orbital motion will reach the ground state where the magnetic interactions will stabilize the electrons further approach toward the proton to remain in its ground state.
All the HA’s in the ground state will not lose any further energy because they all radiate the same SW, so there will not be any further loss of energy. Besides, the interacting magnetic fields also stop any further loss of energy.

The rest is history. Planck eliminated the nature of light as a SW and showed that electrons radiate light only in pulses known as ‘quanta’. Thus, we are only sensitive to this type of light. The, Bohr theory of the planetary HA atom that his mentor suggested was born and is valid.

This to me is basic physics in action.

I do not need to know why the spatial 21 cm radiation is at that level.
I am only concerned about the state of the universe.

The big bang has no direct evidence for its concept except the Hubble Doppler observations.
They modified this to the EoS, which is not valid because Doppler deals with relative motions which has nothing to do with space. The Doppler observations portray us as being in the center of an expanding (explosion?) universe which is a virtual impossibility. It had to be replaced.

The Schroedinger orbitals are shown as clouds surrounding the protons. What happened to the electron? Did it vaporize into this cloud? This is ludicrous. The electron is a particle and a particle it will remain. It also does not disappear during an orbital jump when radiating a photon. It transits as a particle.
Also, the electron is the most important of the two particles. It is the particle involved in all our technology.

You say the electron is not a ‘single point charge’? Then what is the ‘coulomb?

The Ampere illustration shows that this rating is based on the length of a flow of current for ‘one meter’ as I stated. It does not involve a longer flow of current. Limitations have to be made.

CM
Regarding atoms, see above in reply to fortis.

The definition of the ‘spins’ is that they do not spin in the real sense. These terms are used to define that there are reserved positions for one electron in those spin positions.
My definition of spin is that they do have real spin (angular momentum?).

JMB
Thanks for your clarification on classical physics and QM.

__________________
aka Michael Cyrek

cyrek1 wrote
Unfortunately cyrek, thats an oversimplification.
The electron also has wave-like properties and cannot be regarded simply as a classical particle at typical momentums. The use of classical 'paths' to describe electrons is not particularly useful.
Regarding the state transition of bound electrons, the quantum system evolves in quanta only. Even if the electron itself had a time dependent energy evolution, by a linear superposition of states, the atom itself cannot evolve in intermediate 'steps'.

If they are radiating, what are they radiating? The problem with the planetary model is that they would radiate energy in the form of e-m radiation. After all, in your picture you have accelerating charges. This causes problems if you want them to settle down into some sort of ground state.

I'm guessing that your suggestion is that there is some sort of repulsive magnetic interaction between the electron and the proton. (Like two bar magnets aligned with their "norths" pointing in the same direction.) The only snag with this picture is that a lower energy state would exist with the proton's spin being aligned in the opposite direction. The magnetic interaction now becomes attractive, and there is still nothing to stop the electron spiralling down into the proton.
I've already pointed out the problem with this picture, but I should add that there is a conflict between the statement that they will lose no further energy, and the statement that they are radiating (which implies "stuff" going outwards.)

I think that this also shows the advantage of a mathematical theory over a purely "visual" approach. If there were sound theoretical underpinnings to this, then you ought to be able to provide numbers for things like the orbital radius, frequency, ground state energy, etc. that would back up the visula model.

EM radiation may come in quanta (or discrete packets) but the size of ech packet can take on any value. The Schrodinger model of the hydrogen atom is far superior to the planetary model, though it is only an approximation. To deal correctly with electronic spin you need to use the Dirac equation (and even then there are effects that require the use of QFT to pin them down.)

As I've already stated, the planetary model does not explain all of the observed phenomena, such as
But the planetary model that you believe to be correct cannot explain this, whereas QM can. If your model was valid then you would expect to be able to show that this emission occurs with a wavelength of 21 cm. That's another problem with relying on a purely visual model. Without a mathematical basis you can't get very far in making it useful.

Not true (unless you're concerned about possible clustering of redshifts.) An interesting calculation to try is the following. Imagine an explosion with bits being flung out at t=0 with a range of different velocities. To simplify things, make the velocities constant with time. It is obvious that if you sit at the centre of the explosion you will observe that the particle velocities are proportional to distance from the centre (faster particles will travel further after all.) What is perhaps less obvious is what happens if you now position yourself on one of the particles flying out from the centre. Relative to you, you will observe not only that the particles are all flying away from you (with no transverse velocity), but also that the radial velocities are still proportional to distance. Now this isn't the theoretical underpinnings of the BB cosmology, but hopefully it demonstrates that the appearance of a privileged position (e.g. with everything flying away from you with velocities proportional to range) doesn't always mean that you really are in a privileged position.
(This is from an 'A'-level mathematics question from many moons ago. )

You are mis-interpreting what these orbitals are. What you see pictorially represent the wavefunctions corresponding to the eigenfunctions of the Hamiltonian. They indicate the probability density function for finding the electron in a particular region of space. (Or rather, their intensity does.) You can still ask the question "Where is the electron?", but as soon as you try to measure it you collapse wavefunction, and it is no longer in an eigenfunction of the Hamiltonian (or what you may think of as an "energy level").

Let's consider just the ground state (or the 1s orbital), as you talk about the ground state earlier on in your post. The probability density function for the 1s orbital (i.e. the ground state) is spherically symmetric and decays exponentially with distance from the barycentre of atom. You may not be happy with this picture, but it forms the basis for the calculation of the 21 cm wavelength that I pointed out before. (Check out the website that I pointed to.)

First of all, if an electron behaves only as a particle, then how does your model deal with it possessing wavelike characteristics? Secondly, to ask what the electron is doing during a transition is not a meaningful question to ask. This sounds like a cop-out, but it really isn't. We have no reason to believe that the microscopic world should fit in with a world-view based on the macrscopic world. How much experience do you have of observing the world on the atomic scale.

And our technology is based on a QM description of solid state physics. It is QM that is used to model these phenomena, not the planetary model.

I don't say that the electron is not a 'single point charge'. See below.
I will quote from the international definition of the Ampere...
http://physics.nist.gov/cuu/Units/ampere.html
A point charge moving in a circular, planetary orbit, is not at all the same as a uniform current moving along a conductor of infinite length. This is why I compare the electric field of a point charge (which has a 1/r^2 behaviour) with the electric field of an infinite line of charge (which has a 1/r behaviour.) Given this, why would you expect the magnetic field for a moving point charge to be the same as if it were a current moving along an infinite conductor?

If you use the Biot-Savart law, you will see how the fields differ for finite length conductors. You have to use the right equation for the right job.

Thanks Fortis, but let's set aside the enigmatic electon for a moment, because MRI uses protons, and they have a reasonably well-established dimension. Given that, can't we at least approximate a rate of rotation for it, given its charge and magnetic field strength?

I'd like to know this, because it seems a little weird to me that the wavelength equivalent of a proton's rest energy appears to be quite close to a proton's radius. Does that seem even a little fishy to anyone else?

I guess that it was inevitable that I would ramble on about the wrong particle.

In many ways the proton spin is more complicated than the spin of the electron. It is a composite particle consisting of three quarks, each of which possess their own intrinsic spin,contributing to the overall angular momentum of the proton. In addition they appear to contribute orbital angular momentum. The current experimental model has ~30% of the proton spin originating from the spin of the three valence quarks. The rest of the proton spin is thought to originate from orbital angular momentum contributions (which is counter-intuitive, as one might naively have expected the ground state of the particular three quark combination to have zero orbital angular momentum), or possibly from the gluons binding it all together. It looks like the jury's still out on this one.

As you can't consider the proton to be a simple spherical body (for these purposes), a classical analogy may help. Consider the solar system. Here we have a structure with a well defined angular momentum (don't ask me what it is, though, as I haven't a clue ). Like the proton it is also a composite structure consisting of objects with their own intrinsic angular momentum (i.e. rotating planets), as well as orbital angular momentum. Even though this is a well behaved classical system, you would still have great difficulty in assigning a single angular velocity to it.

This has all been a bit rambling (I'm fitting it in between sleeping and calming a baby) but I hope that it helps a bit more than my ramble on the proprties of the electron.

cyrek reply

quanta wrote
Regarding the state transition of bound electrons, the quantum system evolves in quanta only. Even if the electron itself had a time dependent energy evolution, by a linear superposition of states, the atom itself cannot evolve in intermediate 'steps'.

reply
The light we see is quanta pulses of light from a couple(?) of wavelengths in the Balmer series.
However, the HA is radiating a standing continuous sign wave in its energy state depending on its temperature environment from the ground state to a higher state.

In outer space, a recombined electron with a proton can radiate a long Lyman wavelenth from the most outer of the atoms regular energy states to the innermost orbit to radiate the longest Lyman wavelength (hypothetical, of course).

The wave nature of the electron is the result of interactions between its forces. The electron remains a particle during its wave like motions.

You have to understand that what I say may appear to be purely imaginary but I use the knowledge about the nature of these particles to determine their possible dynamic probabilities.

I call this applied science. Ha HA.

The nature of the HA is described by the various characteristics such as its electric charge (coulomb), Its ground state orbit (a sub o), its mass, its velocity, its magnetic characteristics and etc.

__________________
aka Michael Cyrek

I have to ask this, but if it is radiating, where does its energy come from?

The transition that I think you're describing generates the shortest wavelength, corresponding to an energy of ~13.6 eV.

Are you saying that the electron physically moves in a wave-like fashion, and can you quantify what these interactions are and how they lead to the behaviour of the electron?

But in your earlier post, by using the definition of the Ampere in the way that you did, you seem to be implying that the electron is a long linear distribution of charge.

I'm not sure how it counts as being applied science as I can't see any comparison between any prediction and the real world. I've already shown you how the standard QM approach predicts the 21 cm emission band (among many other features), and yet your model doesn't seem to be able to make any such claims.

Thanks Fortis! I've rather dreaded getting into the warped world of quarks, but--you've inspired me to delve forth ;)

It seems kinda strange that things seem to spin/orbit faster at smaller and smaller scales. Is there some kind of spin tensor associated with space at small scales, or is it just because forces get stronger at smaller distances? Hmm...