# Maximum Angular Speed(MAS)

Discussion in 'Physics & Math' started by hansda, Apr 12, 2017.

1. ### The GodValued Senior Member

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So that was dereliction of duty on your part. You knew paddoboy was making mistake and insisting despite my bringing this to his knowledge, you should have intervened. (As per forum guidelines its trolling to insist on mistakes).

You or some mod banned me, I have no alert why? You should realize that calling someone 'ignorant' is certainly asking for retaliation, you could have used 'unaware'. You provoke people and when retaliated, you censure them. That is abuse of power too.

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3. ### Q-reeusValued Senior Member

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To butt in with my non-expert 2 cents worth....While it's perfectly obvious from any standard treatment that electronic orbitals split into both azimuthal i.e. orbital angular momentum, and spin angular momentum, how to try and physically justify the non-zero cases of the former is impossible if all you have are standing waves.
It's hard to find a site that doesn't lump standing and stationary waves as the same thing. A meteorological example of the latter are lee or lenticular waves that form over mountain tops/ranges: https://en.wikipedia.org/wiki/Lee_wave
Translate that into analogous atomic case. Unlike strictly standing waves as superposition of counter propagating traveling waves, a stationary wave solution has net flow of charge naturally leading to net angular momentum and magnetic moment.

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7. ### exchemistValued Senior Member

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Yes I think it does. We use the term "orbital" to denote eigenstates an electron can exist in, within the atom, and some of these have an angular momentum due to displacement of the atom around the nucleus. We use "orbital" rather than "orbit" to denote the fact that there is no defined path that the electron takes, but its motion is such as to give an angular momentum just like an orbiting particle. In this case there are no funnies like the gyromagnetic ratio to worry about, so far as I recall.

Orbitals with a non-zero angular momentum have a node (i.e. zero electron density) at the nucleus, whereas s orbitals do not have this node. This is the QM counterpart of the electron staying away from the nucleus when it has orbital motion, versus "falling in" when it has none.

The number of circumferential nodal planes depends on the number of units of angular momentum the electron possesses: zero for none, one for one unit, two for two units etc. So it is quite elegant: a wave that passes once, per revolution around the atom, from +ve phase to -ve and back for one unit of angular momentum, two such phase changes for 2 units (i.e. the 1st harmonic) and so on.

8. ### originTrump is the best argument against a democracy.Valued Senior Member

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Agreed. It is also safe to say there are not alot (none) of planets that orbit a star in a p, d or f configuration.

9. ### originTrump is the best argument against a democracy.Valued Senior Member

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Your reply...
No thanks. I was just curious what you thought. I guess you do not want to answer?

10. ### QuarkHeadRemedial Math StudentValued Senior Member

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Thank you.
So for each eigenstate/orbital there must exist a spectrum of eigenvalues. And if this spectrum is quantized, let's say it is some integer multiple of the reduced Planck constant $\hbar$.

After rooting out some yellowing prehistoric Phys. Chem. text here I see this spectrum is of the form $\frac{n}{2}\hbar$ where $n = 0,1,2,3,....$ which is identical to the spectrum of the simple quantum harmonic oscillator. Is this significant? Is this why one speaks of "angular momentum" for a bound electron?

Without rooting around, I recall that Bohr, using the "planetary model" for the hydrogen atom, proposed that the allowable angular momentum energy eigenvalues for the electron are simply $n \hbar$, which of course differs from the above by a factor of 2.

Could this therefore mean that we are talking about 2 components for this so-called angular momentum - "orbital" and "intrinsic"?

As you see, I am way out of my comfort zone

11. ### exchemistValued Senior Member

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It is the azimuthal quantum number, l, that determines the number of units of orbital angular momentum. n is the principal quantum number. For any value of n, s orbitals (l=0), have zero angular momentum and are spherical, p orbitals (l=1, 2 lobes) - have one unit, d orbitals (l=2, 4 lobes) have 2 units, f orbitals, l=3, 8 lobes, 3 units) and so on.

The thing I find interesting is the s orbitals. These are the only ones for which the electron density does not fall to zero at the nucleus. In effect the motion might be thought of as chiefly one of falling towards and past the nucleus and out the other side, in a sort of harmonic motion, though due to the uncertainty principle there can be transient angular motion that averages to zero, so the shape of the electron cloud is a spherical ball. It won't be SHM since the "restoring force" is due to Coulomb's Law and decreases with separation.

The spacing between principal levels denoted by n in hydrogen (the simplest case) is proportional to 1/n² so the levels converge to a limit, corresponding to the ionisation energy. In the quantum harmonic oscillator the levels are equally spaced, i.e. do not converge to a limit. It is worth noting that in hydrogen, the s, p, d etc orbitals are all degenerate, i.e. have the same energy as each other. So the value of l and hence the amount of angular momentum, does NOT affect the energy level, in hydrogen. (In atoms with more electrons this is no longer true, due to shielding and other effects which vary depending on the orbital shape.)

12. ### The GodValued Senior Member

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Subject to n being same.

13. ### Q-reeusValued Senior Member

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Try a search using 'Michał Gryziński - free-fall model'. He was no uneducated crackpot, and claimed his model is much more accurate than conventional QM at predicting various scattering phenomena in particular. I make no endorsement, just throw it in for others consideration, or maybe shoot-down based on reliable critiques.

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14. ### hansdaValued Senior Member

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Tell me whom should I follow. Some individual or some accepted well documented text.

Say angular momentum of a particle can be measured only through its magnetic moment because of difficulty in measuring angular momentum directly. So this measurement will give an idea of total angular momentum of a particle.

Consider, if our earth has some net electrical charge. So it will be having a magnetic field around it. This magnetic field will be due 1) orbital angular speed, 2) spin around earth axis and also due precession of the earth.

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Indeed.

16. ### exchemistValued Senior Member

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That's very interesting - I had never come across it. Thanks. Certainly my own picture of electron motion in s orbitals is very like that.

17. ### rpennerFully WiredStaff Member

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Not completely, as there is some fine structure and hyperfine structure at an energy scale about 5-7 orders of magnitude less than the Rydberg energy levels. This is of limited practical importance in visible spectroscopy because hydrogen is a very light atom and so its spectral lines are broadened by Doppler shifts due to thermal motion enough to obscure very fine features except at low temperatures.

http://www.physics.udel.edu/~msafrono/626/Lecture 2-3.pdf

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18. ### exchemistValued Senior Member

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Yes, fair enough. Thanks for clarifying.

19. ### QuarkHeadRemedial Math StudentValued Senior Member

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So I guess I fell at the first hurdle. What do you mean by "number of units of angular momentum"?
If you agree with my recollection of the original Bohr conjecture, namely allowable angular momenta (for an electron in the hydrogen atom) are $\frac{1}{2}n \hbar$, I was assuming that $n$ was an arbitrary non-negative integer. You seem be correcting me by saying that it is the principal quantum number. Is that what you intended?

I am sorry to be dim-witted, but I am unfamiliar with the language here.

20. ### rpennerFully WiredStaff Member

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Quantum numbers of the state of the hydrogen atom:
• n (1, 2, 3, 4, ...) — This is the shell number / "principle quantum number" which determines to first order how much energy the electron has over the ground state: n=1 . This gives the gross structure of the hydrogen emission spectrum.
• (0, ..., n-1) — This is the "orbital quantum number" which determines the total orbital angular momentum of the electron about the proton, |L| = √(ℓ²+ℓ) h/(2π). This has a bigger effect in multielectron atoms and assigned whimsical names which were later reduced to a pattern of letters: (ℓ=0: s, ℓ=1: p, ℓ=2: d, ℓ=3: f, ℓ=4: g, ℓ=5: h, ...)
• m (or $m_{\ell}$) (−ℓ, 1−ℓ, ..., 0, 1, 2, ..., ℓ−1, ℓ) — because quantum mechanics does not allow us to measure every component of angular momentum, we may only know it well in one arbitrary direction. This "magnetic quantum number" which is the projection of L in our chosen direction of measurement. That is typically taken as the z-direction. $L_z = m_{\ell} h / ( 2 \pi )$. This is important in figuring out how a weak magnetic field will split the emission spectrum.
• s ( −½, +½) — because the electron has its own intrinsic angular momentum this "spin quantum number" describes the component of intrinsic angular momentum in the measured direction: $S_z = s h / ( 2 \pi ) = \pm h / ( 4 \pi )$
• I ( −½, +½) — this describes the component of intrinsic angular momentum of the nucleus in the measured direction. For different nuclei, it will be different.
Because of magnetic interactions between the electron and its orbiting motion, the angular momentum of spin and the orbit are not conserved separately. So we need J which is L + S, with quantum numbers j and $m_j$ which replace ℓ, m, and s. Likewise F = I + J, which is important in hyperfine splitting which gives us the 21-cm hydrogen microwave signature used to map the galaxy and the cesium line responsible for the definition of the second.
• j (1/2, ..., n−1/2) — total angular momentum quantum number: $| J | = \sqrt{j (j+1)} \frac{h}{2 \pi}$ .
• $m_j$ ( −j, −j+1, ..., −1/2, 1/2, ..., j−1, j) — analogue of magnetic quantum number. $J_z = j \frac{h}{2 \pi}$ .
It might look like you can add s to ℓ and $m_{\ell}$ to get j and $m_j$ but that's a deceit because these are really labels for eigenstates of the physical systems and the nℓm states of the Schrödinger solution don't correspond simply to the njm states of the Dirac solution. The Dirac solution does not include effects from quantum electrodynamics like the Lamb shift.

https://en.wikipedia.org/wiki/Hydrogen-like_atom has a summary which pretty much requires going to quantum mechanics textbooks to cover better. Because of the variety of names, notation, and pedagogical motivations behind introducing various approximation schemes, it's difficult to find a nice summary.

http://hyperphysics.phy-astr.gsu.edu/hbase/qunoh.html#c2 has an overview of how n, l and m go into solutions of the Schrödinger equation.

So in all cases, the measured component of orbital angular momentum is an integer multiple of $\hbar = \frac{h}{2 \pi}$ (this is hard-coded into the math of quantum mechanics) while the measured component of intrinsic angular momentum of any electron, proton or electron is $\pm \frac{1}{2} \hbar$. Thus we say that these particles have spin-1/2 since $\hbar$ is the scale of quantum angular momentum measurements. The total angular momentum is the square-root of the sum of the squares of angular momentum component on three perpendicular axes which is for an integer, ℓ, (or possible half-integral, j) given as $\sqrt{\ell (\ell + 1) } \hbar$. Because of quantum physics we can't know $L_x$ or $L_y$ if we know $L_z$ but we can know both one component, $L_z$ and the total, $|L| = \sqrt{L_x^2 + L_y^2 + L_z^2}$.

Last edited: Apr 24, 2017
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21. ### exchemistValued Senior Member

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Maybe rpenner's summary answers your questions, but if not I'd be happy to try to explain in my own, somewhat less rigorous, and more pictorial, style.

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22. ### QuarkHeadRemedial Math StudentValued Senior Member

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I thank you both for your trouble. This is a hard subject for me, gimme a day or 2