# Maximum Angular Speed(MAS)

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

1. ### Q-reeusValued Senior Member

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Ohanian's approach works for the photon and electron because there is a correspondence between a Poynting field momentum density ~ ExB applied to circular wave motion components, and the observed spin values. Is there any notion of an equivalent 'E' AND 'B' for a neutrino field? Without a 'magnetic' component, there is no correspondence to the photon or electron cases possible. Additionally very hard to see working from an initial classical basis how the three different varieties of neutrino, with vastly different rest masses, could all wind up with the same same projected magnitude of spin = 1/2.

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3. ### rpennerFully WiredValued Senior Member

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I question that. For one he is tackling classical electromagnetism, so I would put scare quotes around "works." Also, a real-valued, circularly polarized plane wave does not decompose the way you suggest. Calculations provided upon request.

Irrelevant since part III does different operations with the electron spinorial wave.

Presumably one would just sum over the various Dirac fields, one for each quark or lepton.

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5. ### Q-reeusValued Senior Member

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'Components' merely referred to an implied Cartesian basis where if z is propagation axis, you need both x and y components to describe circular polarization of the fields. Ohanian certainly demonstrated the ability to separate into orbital and 'spin' contributions via use of vector identities , and come up with results that match observations, at least to first order accuracy in say electron g-factor.
i.e. go straight for QM basis. Basically highlighting my point. Given the almost totally non-interacting nature of neutrinos, it's clear there can be no sense of any spatially extensive field with appreciable energy-momentum density, unlike for the electron or photon.

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7. ### rpennerFully WiredValued Senior Member

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Yes, that leading term is a consequence of Lorentz symmetry and since the Dirac equation obeys Lorentz symmetry, I though that was given.

One issue is that if you take the real-valued plane wave solution of circularly polarized light in the z direction:
$\phi = 0, \vec{A} = - \frac{E_0}{\omega} \left( \sin ( \omega t - \omega z/c) , \cos ( \omega t - \omega z/c) , 0\right) ; \\ \vec{G} = \frac{1}{\mu_0 c^2} \vec{E} \times \vec{B} = \frac{E_0^2}{\mu_0 c^3} \hat{z}$
(G would be a momentum density)
and apply Ohanian's eqn 3 you find that you have only an orbital angular momentum density:
$| \vec{L} | = \frac{E_0^2}{\mu_0 c^3} \sqrt{x^2 + y^2}$
which gets ( with nothing left over to be spin angular momentum )
while if you apply Ohanian's eqn 4
$\vec{S} = \frac{1}{\mu_0 c^2} \vec{E} \times \vec{A} = - \frac{E_0^2}{\mu_0 \omega c^2} \hat{z} = - \frac{c}{\omega} \vec{G}$
which gives the expected result in light of assuming light is quantisized: $\int \vec{G} dx^3 = \hbar \frac{\omega}{c} \hat{z} \quad \Rightarrow \quad \left| \int \vec{S} dx^3 \right| = \hbar$

So something happened between 3 and 4 where I think Ohanian turned 0 into a spin contribution and a balancing surface term which is wished away, but that assumption doesn't apply to the real plane wave I provide.

In light of QM, what Ohanian is saying that if an electron/neutrino/photon has a not-completely-localized probability amplitude, then one may say the angular momentum is not-completely-localized (i.e. there exists an angular momentum operator and the last step of integrating over all space can be skipped turning it into a (possibly meaningless) denisity) and that angular momentum density can be factored into orbital and intrinsic components.

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8. ### hansdaValued Senior Member

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Does it mean that, spin of a particle has only "intrinsic angular momentum" but no angular speed or rpm?

This concept of Maximum Angular Speed(MAS) may not be very useful with the quantum particles but I see it as a useful concept with astronomical objects. http://newatlas.com/supermassive-black-hole-spin-nustar-newton/26543/

9. ### exchemistValued Senior Member

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Yes that's right, with a QM particle nobody can tell how how "fast" it is "rotating", since we don't have its dimensions (and it has none, if it is a point particle). It is a moot point as to whether there is really anything rotating at all, or at any rate rotating as we know it, Jim. All we can observe is intrinsic angular momentum, via things such as intrinsic magnetic moment.

Yes you may be right in respect of astronomical objects, but I'll have to leave that to the astronomers.

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10. ### Q-reeusValued Senior Member

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It's well known the classical EM result for angular momentum density of a truly plane circularly polarized wave is precisely zero. Only by taking account of the finiteness of any real source can any angular momentum in the wave itself be found. That a torque is induced in an interacting medium impinged by a truly plane circularly polarized wave is a consequence of the finite size of such a medium whose boundary enforces current/charge distributions that do induce net angular momentum:
http://www.opticsinfobase.org/abstract.cfm?URI=oe-13-14-5315
https://arxiv.org/abs/physics/0504082 (and various allied articles by same author at https://arxiv.org/find/physics/1/au: Stewart_A/0/1/0/all/0/1)
(Reminiscent of the need for edge currents to explain the classical prediction of zero susceptibility in a free-electron model conducting solid.)

In the 'particle' case under consideration here, a classical attempt at accounting for photon quantization automatically enforces finite effective spatial extent of E And B fields, and being source free, both must form closed loops. In that sense Ohanian's modelling is 'natural'. Came across a considerably later article by Andre Gsponer that tidies up some loose ends in Ohanian's attempt:
http://arxiv.org/abs/physics/0308027
Latter author never mentions neutrino, and I can't recall Ohanian doing so either (word search not available in that article), which omission is imo telling.

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11. ### danshawenValued Senior Member

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My neutrino physicist friend told me that constructing a neutrino (which has a small amount of mass, and interacts weakly with electroweak bosons and the Higgs to get it) from photons (which have no mass, but which have spin) would break every conservation law used to construct the Standard Model. I agree, it certainly would.

Except for the most important conservation law of all, and that one would be Einstein's.

Spin zero has to mean something tangible or physical. The universe as a unit, and the Higgs field doesn't spin. Particles moving around in it are all made of energy, or excitations of some quantum entangled field like it. Something fundamental is still missing about spin. Many of you here can already taste it.

There isn't a maximum angular speed I know anything about, but there is a minimum, and a particle of spin zero would be it. It is as fundamental an idea about spin as the rest frame is to special relativity. Even something that propagates at the speed of light must do so relative to something that isn't moving, and that would be the rest frame. Without an equivalent rest frame for quantum spin, integer or fractional spin values mean exactly nothing at all physically.

12. ### danshawenValued Senior Member

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Or to put it another way, if something has a quantum spin of +1, and something else of the same mass / energy has a quantum spin of -1, are their relative spins limited by c? I don't think they are, and the double slit experiment with entanglement is why this is the case. Even a tiny angular difference represented by the slit separation is sufficient to observe instantaneous entanglement differences.

13. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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I must be misunderstanding you. We know that light propagates at c independent of the motion of the source or observer and that seems to directly refute your statement. This isn't theory it is direct measurements.

Additionally, what do you mean by the statement, "something that isn't moving?" What would be this thing that isn't moving and how would you possibly know?

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14. ### danshawenValued Senior Member

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I mean that the speed of light, a Lorentz invariant, is always measured with respect to a rest frame. This can be a frame that is in an arbitrary state of relative inertial motion of course. But without a given rest frame, the idea of the speed of light itself is meaningless.

Photons have inertia in a single direction. Particles with inertial mass have inertia in all directions, literally at once, and also literally, faster than light can propagate between them.

15. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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I doesn't seem like you should post in the science sections.

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As you wish.

17. ### hansdaValued Senior Member

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18. ### exchemistValued Senior Member

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Yes indeed, it says exactly what we have been saying, regarding QM "spin".

But it would be interesting to hear from the astronomers and physicists about what happens if a macroscopic object spins with its equator rotating at a significant fraction of the speed of light (relative to the axis of rotation).

19. ### Q-reeusValued Senior Member

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https://arxiv.org/abs/1408.4145v1
The kind of golly gee figures an ex member liked to post often.

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20. ### hansdaValued Senior Member

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What I understand is that spin of a QM particle has two components. One component is due orbital angular speed, as our Earth orbits the Sun. The other component is due rotation around self axis, as the earth spins around its own axis. In addition the particle may also precess, as our Earth is also precessing.

This will cause rotational frame-dragging.

21. ### originIn a democracy you deserve the leaders you elect.Valued Senior Member

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So, did you not read the responses or do you just disagree with them?

22. ### exchemistValued Senior Member

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Not exactly. With a QM entity, what is referred to as "spin" is the intrinsic angular momentum ONLY. Angular momentum due to displacement of the particle is not called spin. In at atom, an electron may have orbital angular momentum as well as intrinsic angular momentum. But equally it may not: for example, in s orbitals, there is zero orbital angular momentum.

23. ### QuarkHeadRemedial Math StudentValued Senior Member

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Quick question, exchemist - since the abandoning of the planetary model of the atom, does it still make sense to talk about "orbital angular momentum" of the electron?