[hamdani yusuf (OP)]

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

In macroscopic objects, electrically charged particles interact according to Coulomb's law. Is it still working in microscopic objects?
How does it contribute to the formation of atomic orbitals?

The experiment below shows that mass of atomic nucleus affects the emission spectrum, thus it also affects atomic orbitals. Is there any known formula relating them?
http://myslu.stlawu.edu/~jmil/physics/legacy/student_projects/2001/fiacco.shtml

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

[paul cotter]

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.

[hamdani yusuf (OP)]

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.

[alancalverd]

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?

[Bored chemist]

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?

Actually, slightly less.
They describe how it travels through a homogeneous medium.
They don't, for example, explain the blue sky.

[hamdani yusuf (OP)]

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.

[hamdani yusuf (OP)]

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.

[alancalverd]

They don't, for example, explain the blue sky.

Rayleigh scattering in the atmosphere is a classical continuum effect, entirely consistent with Maxwell.

[Bored chemist]

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.
In which case, the medium has to be "lumpy", but Maxwell's eqns don't deal with lumps.
They are consistent, but they won't model it.

[alancalverd]

Nobody said they do. The lift equation doesn't model the ground run correction, but they are entirely consistent with the business of taking off and landing because it's all classical continuum mechanics.

[Bored chemist]

Nobody said they do.

I'm sure someone did...

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.

[alancalverd]

Once the scattering has occurred, the spectrally-shifted radiation propagates exactly according to Maxwell because it is electromagnetic radiation.

I seem to have to keep repeating the bloody obvious: Maxwell's equations describe the propagation of EM radiation, not its interactions nor even its generation (the generation of radio waves is easily explained with them but atomic spectra and x-rays need a different approach - quantum mechanics is helpful) .

[Bored chemist]

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.
And nature abhors a vacuum.
So, exactly where do they apply?
Before you answer that, just pause and consider how big an atom is.

[alancalverd]

Maxwell's equations only describe the propagation of EM radiation in a vacuum.

No. They apply to any medium if you substitute  ε_m and μ_m for ε₀ and μ₀. The vacuum value is admittedly an experimental approximation, but none the worse for that.

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.

[Bored chemist]

It's like swimming in treacle...
Please show me the bit of Maxwell's equations that says "sometimes, some of the light shoots out sideways".
For scattering you need two different values of ε and μ- one of each of the medium that's scattering and one of each for the matrix in which it's suspended.
For example, the air molecules have different values from the "vacuum" which is "between" them.

Please show me which of Maxwells' eqns has two values of μ and ε and a value for the size of the particle.

[Eternal Student]

Hi.

No. They apply to any medium if you substitute  ε_m and μ_m for ε₀ and μ₀.

   I would more or less agree with that.   Let's do the "more" bit first:
You can exhibit the propagation of an e-m wave through a simple material with a speed 1/√(μ_mε_m) < c    when you reformulate Maxwells equation by replacing the E and B fields with D and H fields where necessary.    For a simple dielectric material,  the relationship between the electric polarisation, P, of the material and an underlying E field should be linear (so you will have the constitutive relationship  D  =   ε_m  E etc.).   Similarly for a simple magnetic material the magnetic susceptibility of the material should also remain constant so that you have a linear constitutive relationship B = μ_m H between the B and H fields.   
    Overall then,  what you ( @alancalverd ) are saying is that provided you consider "Maxwells equations" to be the version which is often called "Maxwells equations in matter"  rather than "Maxwells equations in a vacuum" then everything works fine and you can exhibit a wave that propagates as required with an appropriate speed for the medium.    I would agree with that - for simple linear materials.

So that brings us to the "less" bit:

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.

    1.  Hard Ferromagnetic material can retain a Magnetisation even when the H field is reduced back to 0 after first being a strong non-zero field in some direction.  So the relationship between  B and H fields inside the material is not the simple linear relationship we would want,  instead it can depend on the history of the fields the material has been exposed to.

    2.  Superconducting materials also have complicated relationships between B and H fields:   For example, in Type-I superconductors we can have Magnetic susceptibility χ_m = -1 throughout some critical range of the H field (giving a relative permeability μ_r =  1 + χ_m = 0 ) but then undergoes a discontinuity and we have  χ_m = 0   (and hence μ_r immediately changing to 1)  outside that range.

    3.   Similarly, not all dielectric materials will be simple linear materials (where the D and E fields would be linearly related).   I've not studied it but I have been informed that sometimes the dielectric polarisation of a material isn't even in the same direction as the applied E field (e.g. in some crystalline structures we require a rank-2 tensor, T, to relate the (vector) E field to the (vector) D field:   D  = T E     )

    I would not vouch for how (or even "if") an e-m wave can propagate through some of these non-linear materials.   For all I know, the propagation of an e-m pulse through some non-linear materials could be extremely unusual:
1.  A pulse of e-m radiation sent into the material could travel through the material along path(s) that may not be straight lines.
2.  It may not always take the same path but instead may depend on the prior history of E and B fields that were applied to that material.   Since the e-m wave is itself changing the E and B fields inside the material when it passes through, the first part of the pulse may exit the material in a different place to where later parts of the pulse exit the material.

Best Wishes.

[Bored chemist]

If you get a high enough field strength practically everything behaves non-linearly.
Even air  misbehaves if you try hard enough.

[hamdani yusuf (OP)]

If you get a high enough field strength practically everything behaves non-linearly.
Even air  misbehaves if you try hard enough.

I'm wondering how it would look like under different air pressure.

[hamdani yusuf (OP)]

Do Maxwell equations explain electrostatic and magnetostatic interactions?
If they don't, what does?
Is it compatible with Maxwell equations?

[Bored chemist]

If you get a high enough field strength practically everything behaves non-linearly.
Even air  misbehaves if you try hard enough.

I'm wondering how it would look like under different air pressure.

Something like this will give you a clue.
http://lamptech.co.uk/Documents/M2%20Pressure.htm
If you get the pressure high enough you will blur out the spectral lines and et what looks like BBR.