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Originally Posted by cyrek1
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.
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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.
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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.
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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.
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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.
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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.
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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.
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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
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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.
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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.
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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.
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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.
)
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The Schroedinger orbitals are shown as clouds surrounding the protons. What happened to the electron? Did it vaporize into this cloud? This is ludicrous.
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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.)
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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.
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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.
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Also, the electron is the most important of the two particles. It is the particle involved in all our technology.
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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.
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You say the electron is not a ‘single point charge’? Then what is the ‘coulomb?
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I don't say that the electron is not a 'single point charge'. See below.
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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.
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I will quote from the international definition of the Ampere...
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The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2 x 10-7 newton per meter of length.
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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.