Rotation

Outline

Ideas about rotation are intimately related to notions of motion. Talk of space is all over the place. But, by the premise of EthnoPhysics, we cannot start an analysis of rotation with hand-waving bluster about space. Rather we insist on discussing sensation, namely achromatic visual sensation – greyness – as a key experience for understanding rotation. In everyday speech, a description of rotation expresses how a particle moves about in some repetitive way. We often use words like spiraling, gyrating or looping to describe rotary motion. But for a scientific discussion, we are going to develop the term angular-momentum.

Variations in brightness, of daybreak and nightfall, are how we initially know our world is turning. So an analysis of visual sensation, with little reference to space, is how we start talking about angular-momentum. This sort of rotation is marked by $\mathrm{\overline{S}}$ and associated with the word spin. Then later, as spatial concepts are introduced, we develop the frame dependent aspect known as orbital angular-momentum. But before discussing all that, we first consider an even simpler way to describe rotating particles, their greyness.

Greyness

To mathematically describe greyness, recall that EthnoPhysics started with binary descriptions that crudely divided all achromatic visual perception into just four possibilities; a black or white sensation in either the left or right eye. These four sensations were objectified to define up quarks and down quarks noted by $\mathsf{u} , \: \mathsf{\overline{u}} , \: \mathsf{d} \, \text{ \sf{and} } \, \mathsf{\overline{d}} .$ Quarks were counted to determine four quark coefficients $n^{\mathsf{u}},$ $n^{\mathsf{\overline{u}}},$ $n^{\mathsf{d}}$ and $n^{\mathsf{\overline{d}}} .$ And quark-coefficients were used to define two seed coefficients as

$N^{\mathsf{U}} \equiv n^{\overline{\mathsf{u}}} + n^{\mathsf{u}}$

and

$N^{\mathsf{D}} \equiv n^{\overline{\mathsf{d}}} + n^{\mathsf{d}}$

Then rotation is characterized by the greyness quantum-number $G$ which is defined by

$G \equiv \dfrac{ \; N^{\mathsf{U}} - N^{\mathsf{D}} }{8}$

Sensory interpretation: Up-quarks and down-quarks are objectified from achromatic visual sensations. So if $G \! > \! 0$ then white sensations outnumber black sensations. Altogether, they look bright or light-grey. For $G \! < \! 0$ black sensations are more numerous than white sensations. Overall, they look dark-grey. So the greyness indicates whether a complex visual experience is collectively bright or dark. If $G \! = \! 0$ then we say that P is just medium grey.

Greyness is related to the total angular-momentum quantum-number by $\textsl{\textsf{J}} = | \hspace{0.5px} G \hspace{0.5px} | .$ It is also closely related to another way of describing rotating quarks, their chirality.

Chirality

To specify the chirality, recall that achromatic visual sensations are given a binary description relative to the reference sensation of seeing the Sun. Perceptions are objectified as either up quarks or down quarks that are noted by $\mathsf{u} , \: \mathsf{\overline{u}} , \: \mathsf{d} \, \text{ \sf{or} } \, \mathsf{\overline{d}} .$ Both up and down quarks are classified together as rotating quarks. Now let some particle P be described by $N^{\mathsf{U}}$ and $N^{\mathsf{D}} ,$ sums of quark coefficients where

$\Delta n^{\mathsf{U}} \equiv n^{\overline{\mathsf{u}}} - n^{\mathsf{u}}$

and

$\Delta n^{\mathsf{D}} \equiv n^{\overline{\mathsf{d}}} - n^{\mathsf{d}}$

Then the rotation of P is characterized by a number called the chirality which is noted by $\delta_{\mathsf{o}}$ and defined as

$\delta_{\mathsf{o}} \equiv \begin{cases} +1 & \mathsf{\text{if}} \; \; \Delta n^{\mathsf{U}} < \Delta n^{\mathsf{D}} \\ \; \; 0 & \mathsf{\text{if}} \; \; \Delta n^{\mathsf{U}} = \Delta n^{\mathsf{D}} \\ -1 & \mathsf{\text{if}} \; \; \Delta n^{\mathsf{U}} > \Delta n^{\mathsf{D}} \end{cases}$

If $\delta_{\mathsf{o}} \! = \! +1$ then P is said to be dextrorotatory. If $\delta_{\mathsf{o}} \! = \! -1$ then P is levorotatory. And if $\delta_{\mathsf{o}} \! = 0$ then we say that P is achiral. Chirality is an intrinsic particle characteristic. It depends only on the quarks contained in P, not on some external reference frame. So a particle that is classified as dextrorotatory, levorotatory or achiral will remain in the same category for any frame of reference.

Now recall that by the anti-commutative property of subtraction any differences in the quark-coefficients of some particle $\mathsf{P} ,$ and its conjugate twin $\overline{\mathsf{P}} ,$ are related as $\Delta n^{\mathsf{Z}} (\mathsf{P}) = - \Delta n^{\mathsf{Z}} (\overline{\mathsf{P}}) \, .$ Then by the multiplicative property of inequality we know that if

$\Delta n^{\mathsf{U}} (\mathsf{P}) \; < \; \Delta n^{\mathsf{D}} (\mathsf{P})$

then

$\Delta n^{\mathsf{U}} (\overline{\mathsf{P}}) \; > \; \Delta n^{\mathsf{D}} (\overline{\mathsf{P}})$

This means that if $\mathsf{P}$ is dextrorotatory, then $\overline{\mathsf{P}}$ is levorotatory, and vice versa. The foregoing definition implies that particles and their conjugate twins always have opposite chirality

$\delta_{\mathsf{o}} ( \mathsf{P} ) = -\delta_{\mathsf{o}} ( \overline{\mathsf{P}} )$

For example, a proton in its ground state contains four down anti-quarks and no up-quarks. So $\Delta n^{\mathsf{U}} \! = \! 0 ,$ $\Delta n^{\mathsf{D}} \! = \! 4$ and the proton is dextrorotatory. However a ground state electron contains four up anti-quarks and no down-quarks. So it is levorotatory. But if this electron absorbs a dark photon containing eight down-quarks, then it becomes dextrorotatory.

Achiral Particles

Consider the chirality $\delta_{\mathsf{o}}$ of some generic particle P. If $\delta_{\mathsf{o}} \! = 0$ then $\Delta n^{ \mathsf{U}} \! = \! \Delta n^{ \mathsf{D}} .$ Or, in terms of individual quark-coefficients $n^{\overline{\mathsf{u}}} \! - n^{\mathsf{u}} = n^{\overline{\mathsf{d}}} \! - n^{\mathsf{d}} .$ When P meets this requirement it is called an achiral particle.

The achiral condition can also be expressed in terms of the greyness $G .$ Recall that a set noted by $\mathbb{D} ,$ which contains just the ordinary-quarks from P, is called its dextral component. Whereas another set $\mathbb{S} ,$ which contains all of P’s anti-quarks, is called its sinistral component. Since these sets are uniformly composed from either all ordinary-quarks or all anti-quarks, their greyness is given by

$G ( \hspace{1px} \mathbb{S} \hspace{1px} ) = \dfrac{ \; n^{\overline{\mathsf{u}}} - n^{\overline{\mathsf{d}}} }{8}$

and

$G ( \hspace{1px} \mathbb{D} \hspace{1px} ) = \dfrac{ \; n^{\mathsf{u}} - n^{\mathsf{d}} }{8}$

So the condition that $\delta_{\mathsf{o}} \! = 0$ implies that $G ( \hspace{1px} \mathbb{S} \hspace{1px} ) \! = \! G ( \hspace{1px} \mathbb{D} \hspace{1px} ) .$ That is, the collective greyness of P’s anti-quarks is the same as for its ordinary quarks. By taking absolute-values, the achiral condition can be expressed in terms of $\textsl{\textsf{J}} \hspace{1px} ,$ the total angular-momentum quantum-number, as

$\textsl{\textsf{J}} \hspace{0.5px} ( \hspace{1px} \mathbb{S} \hspace{1px} ) = \textsl{\textsf{J}} \hspace{0.5px} ( \hspace{1px} \mathbb{D} \hspace{1px} )$

This achiral quality does not change if up and down quarks get confused for each other. The description of black and white could be completely backward, yet an achiral particle would still be classified as achiral. Moreover if some quarks get switched with their anti-quarks, an achiral particle would still be designated as achiral. And furthermore, an achiral particle will remain achiral in any frame of reference.

So achiral particles have an intrinsic quality that is very robust and certain. Up-seeds and down-seeds are distributed equitably between quarks and anti-quarks. Achirality is remarkably independent of references to other standards or external conditions. In terms of sensation, achiral particles are objectified from black and white experiences that are equally balanced between left and right-sides. Their greyness is laterally symmetric. Both eyes see the same level of brightness or darkness. And this trait is fixed, even if black and white get mixed-up, and even if left and right get mixed-up, and even if the reference-frame gets changed. Achirality has an absolute quality.

In everyday speech, describing rotation expresses how a particle moves. But scientifically, motion is always understood relative to a reference frame. So a scientific description of rotation must also be frame dependent. Indeed a relative description of rotation can be based on differences in $\overline{\omega} \hspace{1px} ,$ the intrinsic angular-velocity.

The relative angular-velocity of a material particle P, in a frame-of-reference F, is defined by

$\overline{\Omega}^{\, \mathsf{P}}_{\, \mathsf{F}} \, \equiv \, \overline{\omega} \left( \mathsf{P} \right) \hspace{-0.5pt} - \, \overline{\omega} \left( \mathsf{F} \right)$

Then if $\overline{\Omega} \! = \! \left( 0, 0, 0 \right)$ we say that P is not rotating in the F-frame. Otherwise, we say that P is rotating in the F-frame. This is conceptually straightforward. But actually determining $\overline{\Omega}$ depends on specific details like particle positions and the alignment of axes. It can be quite complicated.

However calculations can be simplified by choosing reference-frames that are achiral or which have a negligible angular-speed. Recall that the intrinsic angular-velocity $\overline{\omega}$ has been defined by a product of the chirality with an angular-speed. So for these chosen frames, $\overline{\omega} \! \left( \mathsf{F} \right)$ is nil.

In practice, there are two common ways to obtain a null value for $\overline{\omega} \! \left( \mathsf{F} \right).$ The snapshot method and the Stonehenge technique. First, the snapshot method where measurements are made so quickly that any changes in F are small enough to be ignored. Then the frame’s angular-speed is supposedly negligible and therefore $\overline{\omega} \! \left( \mathsf{F} \right)$ is nil. For example this technique has been highly refined in Toronto where ultra-fast femtosecond laser pulses have been used to investigate water molecules.¹M.B. de Kock, S. Azim, G.H. Kassier and R.J.D. Miller, Determining the radial distribution function of water using electron scattering. Journal of Chemical Physics, Volume 153, Issue 19, 2020.

And second, consider the Stonehenge method where observations are made over long periods of time, perhaps many years or centuries. Like Stonehenge the frame F must be permanently attached to a fixed location on Earth. Then any difference between the number of days and the number of nights at that location is small enough to be ignored. That is, we can assume that F provides a background that includes an enormous quantity of bright and dark experiences, in equal amounts. These black and white visual sensations are presumably evenly spread over all views, angles and sides. They are uniformly distributed, and so the greyness of F is supposedly homogeneous. Then the frame’s chirality $\delta_{\mathsf{o}} \! \left( \mathsf{F} \right)$ is zero and therefore $\overline{\omega} \! \left( \mathsf{F} \right)$ is nil.

These two approaches are not mutually exclusive. We often employ both methods by briskly doing experiments in laboratories that are firmly attached to the Earth. The combination is usually enough to ensure that F effectively has no angular-velocity. Then the relative description of rotation collapses into a simple account of P’s intrinsic character. For many common situations

$\overline{\omega} \! \left( \mathsf{F} \right) = \left( 0, \, 0, \, 0 \right)$

so that

$\overline{\Omega}^{\, \mathsf{P}}_{\, \mathsf{F}} = \hspace{1px} \overline{\omega} \! \left( \mathsf{P} \right)$

This hides the way that scientific descriptions of rotation actually depend on having a reference frame. Relativity is thus veiled.

So unless F is something like a roller-coaster or a merry-go-round, we can casually say that P is not rotating if $\overline{\omega} \! \left( \mathsf{P} \right) \! = \! \left( 0, 0, 0 \right) .$ And that it is rotating if $\overline{\omega} \! \left( \mathsf{P} \right) \! \ne \! \left( 0, 0, 0 \right) .$

Furthermore, recall that $\omega \propto \! \sqrt{\textsl{\textsf{J}} \; \, }$ where $\textsl{\textsf{J}}$ is the total angular momentum quantum-number of P. So usually we can say that P is not rotating if $\textsl{\textsf{J}} \! = \! 0$ and that it is rotating if $\textsl{\textsf{J}} \! \ne \! 0 \, .$

Rotating Frames

Any reference frame F is a compound-quark. And so the quarks in F determine its intrinsic character just like they do for other particles. However logical issues arise if we try to define relative characteristics for F. The problem is determining what superseding standards can replace the initial reference standards which were supposed to be all encompassing.

This question was historically relevant for marine navigation where distant stars were used to replace local indicators. The celestial approach was good enough for practical purposes. But the logical issue just shifted to controversy about the ‘distant stars’ which were necessarily far away and therefore inaccessible.

The issue is especially confusing in discussions about rotation because many common reference-frames are at least approximately achiral. And this obscures the principle that scientific descriptions of motion are only understood relative to a frame of reference. The muddle has led to several hundred years of inconclusive analysis about rotating buckets.²Isaac Newton, Mathematical Principles of Natural Philosophy, page 10. Translated by Andrew Motte and Florian Cajori. University of California Press, 1946.^, ³Ernst Mach, The Science of Mechanics, second edition page 231. Translated by Thomas J. McCormack. The Open Court Publishing Company, Chicago 1902. And also to many experimentally inaccessible assertions concerning Mach’s principle. Anyway, questions about rotating reference-frames persist to this day.

For EthnoPhysics, the issue is viewed as follows: Rotation is a kind of motion. And since Einstein, we understand that motion is always relative to a reference-frame. So we consider the idea of a ‘rotating reference-frame‘ to be logically inconsistent, a contradiction in terms. The proposition is just scientific nonsense unless another superseding reference-frame is immediately identified. Nonetheless, the chirality of a reference-frame is well-defined. Therefore an analysis of chirality is our preferred way of scientifically grappling with the dizzy concept of a ‘rotating frame‘.

So instead of mumbling about ‘non-rotating frames‘ we would rather talk about achiral frames. For EthnoPhysics rotation is relative, but achirality is absolute.

A Whirl of Excitement

Next we push the discussion of rotating-quarks to develop the idea of rotation more broadly. Recall that achromatic visual sensations are given a binary description relative to the reference sensation of seeing the Sun. Perceptions are objectified as either up quarks or down quarks that are noted by $\mathsf{u} , \: \mathsf{\overline{u}} , \: \mathsf{d} \, \text{ \sf{or} } \, \mathsf{\overline{d}} .$ Counting these quarks then gives the total number of up or down quarks as $N^{\mathsf{U}}$ or $N^{\mathsf{D}} .$

Revolving and rotating are ways that a particle can move. And we often refer to a moving particle as being excited. So twirling and revolving are also considered to be modes of excitation. Movement is related to interactions between particles. So for example particles that have been excited by the absorption of many photons usually contain many quarks. Thus we can make a crude description of the magnitude or degree of excitation, just by counting quarks.

Here is one way of describing a particle’s level of excitation. It depends on adding-up the total number of all rotating quarks. So it is called the rotary quantum number and defined by

$\widehat{\mathrm{n}} \equiv \dfrac{ \; N^{\mathsf{U}} + N^{\mathsf{D}} }{8}$

Other ways of describing excitation depend on counting other kinds of quarks. For example, the principal quantum number is defined as

$\mathrm{n} \equiv \dfrac{ \; n^{\mathsf{d}} \; }{4}$

where $n^{\mathsf{d}}$ is a quark coefficient for ordinary down-quarks. We discuss the principal quantum number $\mathrm{n}$ in much more detail later because it is widely used for understanding atomic spectra. But before that, we examine pointy particles.

The norm is used to describe spatial relationships, perhaps somewhat as suggested by the patterns in this ajat basket from Borneo. — Ajat basket, Penan people. Borneo 20th century, 23 (cm) diameter by 36 (cm) height. Hornbill motif. Photograph by D Dunlop.

Pointy Particles

Consider some profoundly plain particles. The idea here is that these particles have no distinct orientation in space. No variations in length or width. No bumps or edges that could be related to an external reference frame. And let them also be very small. So small that they are difficult to detect. Then these plain particles approximate the size and simplicity of a geometric point. But we will have to discuss space before we can make a good definition of a geometric point. So for now we just call these little quanta pointy particles.

Here is a definitive list of pointy particles: Seeds, quarks, nuclear-particles, atomic nuclei, and multi-electron atoms with full shells. For EthnoPhysics, all these particles are defined from sense perceptions. Thus a class of pointy particles is also defined from sensation.

Spin

In physics, some words like ‘reality’ and ‘strange’ are overworked. They become so tired that their meaning gets clouded by mental shortcuts and fuzzy clichés. We try to avoid them, but unfortunately ‘spin’ is one of those words. Nonetheless, the ideas associated with spin are important. So in the following discussion we restrict usage to just the names of specific, well-defined quantities. And thereafter, we shun casual use of the term.

Spin Quantum-Number

Particle rotation is evaluated using a number $s ,$ called the spin quantum-number. For pointy particles $s$ has exactly the same value as the total angular momentum quantum-number $\textsl{\textsf{J}} \, .$ Recall that $\textsl{\textsf{J}}$ is defined from a difference in the coefficients of rotating quarks. So for pointy particles, $s$ is given by

$s = \textsl{\textsf{J}} \equiv \dfrac{ \, \left| \, N^{\mathsf{U}} - N^{\mathsf{D}} \, \rule{0px}{10px} \right| \, }{8}$

In general, if a particle is not pointy, then $s \! \ne \! \textsl{\textsf{J}} \, .$ We talk about non-pointy particles later when considering orbital angular momentum. But nuclear particles are all pointy, and their quark models show that

$s \in \left\{ 0\, , \; 1 \! /2\, , \; \; 1\, , \; \; 3 \! /2\, , \; \; 2 \, \; \ldots \rule{0px}{10px} \right\}$

If $s$ is an integer, particles are called bosons. If $s$ is half an odd-integer, they are called fermions.

Spin Angular-Momentum Vector

In a quark space centered on particle P we can define a vector $\mathrm{\overline{S}}$ to describe P’s rotation. Let $\widehat{\rho}$ mark the radial direction vector of P and let $s$ note its spin quantum-number. The chirality is written as $\delta_{\mathsf{o}} .$ Then the spin angular-momentum vector $\mathrm{\overline{S}}$ is defined by

$\raisebox{-2px}{\overline{\mathrm{S}} } \equiv \dfrac{\, h \delta_{\mathsf{o}}}{2\pi} \sqrt{s(s+1) \, } \, \widehat{\rho}$

For pointy particles $s$ depends just on the quarks in P. And all nuclear particles are pointy. So for nuclear particles $\mathrm{\overline{S}}$ is an intrinsic characteristic. It does not depend on any frame of reference.

The spin angular-momenta of any particle $\mathsf{P}$ and its conjugate twin $\overline{\mathsf{P}}$ are identical. This is because their radial-direction vectors are related as $\widehat{\rho} ( \mathsf{P} ) \! = \! - \widehat{\rho} ( \mathsf{\overline{P}} ) .$ And as discussed earlier, their chirality is set by $\delta_{\mathsf{o}} ( \mathsf{P} ) \! = \! -\delta_{\mathsf{o}} ( \overline{\mathsf{P}} ) .$ These two negative factors cancel each other so that

$\overline{\mathrm{S}} ( \mathsf{P} ) = \overline{\mathrm{S}} ( \mathsf{\overline{P}} )$

If $\widehat{\rho} \! = \! (0,0,0)$ then the norm of the spin angular-momentum vector is zero. Otherwise

$\left\| \hspace{1px} \raisebox{-2px}{\overline{\mathrm{S}}} \hspace{1px} \rule{0px}{9px} \right\| = \dfrac{h}{2\pi} \sqrt{s(s+1) \, }$

Sensory Interpretation: The vector $\mathrm{\overline{S}}$ describes the distribution of achromatic visual sensations that are objectified in particle P. If $\| \overline{\mathrm{S}} \| \! = \! 0$ then the color of P is a dull middling grey.

Spin-Projection Quantum-Number

Consider some particle P that is described in quark space. Let the central axis in this space be noted by the unit-vector $\widehat{z} \, .$ We can assess P’s spin angular-momentum vector $\mathrm{\overline{S}}$ by its projection onto the central-axis. Thus

${\mathrm{S}}_{z} \, \equiv \, \overline{\mathrm{S}} \cdot \widehat{z}$

This component ${\mathrm{S}}_{z}$ is also represented by $m_{s}$ a number called the spin-projection quantum number defined by

$m_{s} \equiv \dfrac{2\pi}{h} S_{z}$

The numerical values assumed by $m_{s}$ are not continuous because any trait defined from quark-coefficients is quantized. And the smallest possible change in $s$ is presumably due to an interaction with a single photon such that $\Delta s \! = \! \pm 1 .$ So as $\mathrm{\overline{S}}$ varies between being parallel or antiparallel to $\widehat{z} \, ,$ the possible values of $m_{s}$ range over

$m_{s} \in \left\{ \, -s , \; -s \! + \! 1 , \; -s \! + \! 2 \; \; \ldots \; \; s \! - \! 2 , \; \, s \! - \! 1 , \; \, s \, \rule{0px}{10px} \right\}$

The actual value of $m_{s}$ depends on specific arrangements. For solitary protons or electrons $m_{s} \! = \delta_{\mathsf{o}} / 2 \, .$ But for the electron⁴This spin-projection quantum-number for the electron can be written-out in terms of the quark coefficients of the hydrogen atom as $m_{s} = ( n^{\mathsf{u}} + n^{\mathsf{\overline{u}}} - 3 n^{\mathsf{d}} + n^{\mathsf{\overline{d}}} ) / 8 \, .$ in a hydrogen atom $m_{s} \! = \widehat{\mathrm{n}} - 2\mathrm{n} \hspace{1px} .$ Thus ${\mathrm{S}}_{z}$ is closely related to the rotary quantum-number, the principal quantum-number and the spin-projection quantum-number.

Spinors

The vector $\mathrm{\overline{S}}$ has two other components in addition to ${\mathrm{S}}_{z} \, .$ But they remain vague because there are limits to how narrowly we can parse various shades of grey. To assess these limitations, we next review how achromatic visual sensations have been analyzed.

Recall that EthnoPhysics started with binary descriptions that crudely divided all greyish perception into just four possibilities; a black or white sensation in either the left or right eye. These four sensations were objectified to define four quarks noted by $\mathsf{u} , \: \mathsf{\overline{u}} , \: \mathsf{d} \, \text{ \sf{and} } \, \mathsf{\overline{d}} .$ Quarks were counted to determine four quark coefficients $n^{\mathsf{u}},$ $n^{\mathsf{\overline{u}}},$ $n^{\mathsf{d}}$ and $n^{\mathsf{\overline{d}}} .$ And then quark-coefficients were used to define two seed coefficients as

$N^{\mathsf{U}} \equiv n^{\mathsf{\overline{u}}} + n^{\mathsf{u}}$

and

$N^{\mathsf{D}} \equiv n^{\mathsf{\overline{d}}} + n^{\mathsf{d}}$

For these seed-coefficients, quarks and anti-quarks are added together in the same way. Lateral sensations have no special significance in the sum. And seed-coefficients thereby discount the distinction between left and right.

Calculations based on $N^{\mathsf{U}}$ and $N^{\mathsf{D}}$ give the same result even if there is some confusion or mix-up between left and right. This indifference is useful because it simplifies calculations and makes them less dependent on specific arrangements. Using seed-coefficients instead of quark-coefficients is a way to transcend sensory detail. It is an important way of objectifying a description.

But analysis based on using just $N^{\mathsf{U}}$ and $N^{\mathsf{D}}$ is less precise. These numbers depend on a binary distinction between black and white. So ultimately, they can offer only two degrees-of-freedom for theoretical consideration.

Recall that the quantum-numbers $\widehat{\mathrm{n}} ,$ $s$ and $\textsl{\textsf{J}}$ have all been defined from $N^{\mathsf{U}}$ and $N^{\mathsf{D}} .$ So descriptions based on these quantum-numbers also have at most two degrees-of-freedom. We can represent various particles using $\| \hspace{0.5px} \overline{\mathrm{S}} \hspace{0.5px} \|$ and ${\mathrm{S}}_{z} .$ But adding more detail about the other two components is superfluous. So all three components have not been defined.

However there are some perfectly good two-dimensional vectors written like $\vert s, m_{s} \rangle .$ And they can be treated as spinors to access results from mathematical group theory. This approach is used for multi-electron atoms and complex nucleii. But it is not necessary for our treatment of atomic hydrogen, and so we go no further with spinors.

Thus overall, our analysis of $\mathrm{\overline{S}}$ offers just two degrees of descriptive freedom. But physical rotation is certainly a three-dimensional phenomenon. So a full description of rotation requires more than just $s$ and $m_{s} \, .$ Something is missing. What could it be?

The Handedness

Next we reconsider the sweet sensations that were associated with a taste of honey. Recall that these pleasant flavours were given a binary description as either sugary or savoury. Sugary sensations were objectified as dextro quarks, $\mathbf{d}$ or $\overline{\mathbf{d}}$ . And savoury sensations were reified as the levo quarks $\mathbf{l}$ or $\overline{\mathbf{l}}$ . Both dextro and levo quarks were classified together as stereochemical quarks.

Oral sensations are almost negligible compared to seeing the Sun. And stereochemical quarks are about a billion times smaller than up-quarks. But nonetheless, oral sensations are still important because they provide the missing perception required for a fully three-dimensional description of rotation. This nuance is called the handedness and defined as follows. Let some particle P be characterized by $\Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}}$ and $\Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{D}}}}} ,$ the coefficients of its levo and dextro quarks. That is

$\Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} \equiv n^{\overline{\mathbf{\,l}}} - n^{\mathbf{\, l}}$

and

$\Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{D}}}}} \equiv n^{\overline{\mathbf{d}}} - n^{\mathbf{d}}$

Then the rotation of P is characterized by a number called the handedness which is noted by $\delta_{\! \concavediamond}$ and defined as

$\delta_{\! \concavediamond} \equiv \begin{cases} +1 & \mathsf{\text{if}} \; \; \Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} < \Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{D}}}}} \\ \; \; 0 & \mathsf{\text{if}} \; \; \Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} = \Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{D}}}}} \\ -1 & \mathsf{\text{if}} \; \; \Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} > \Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{D}}}}} \end{cases}$

If $\delta_{\! \concavediamond} \! = \! +1$ then P is said to be right handed. If $\delta_{\! \concavediamond} \! = \! -1$ then P is left handed. And if $\delta_{\! \concavediamond} \! = 0$ then we say that P has no handedness. Thus rotation is more completely described using stereochemical quarks. But the internal energy of stereochemical quarks is measured in milli electronvolts. Stereochemical quarks are so small that their icons and images are often just omitted from diagrams.

$\Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} (\mathsf{P}) \; < \; \Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{D}}}}} (\mathsf{P})$

then

$\Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{L}}}}} (\overline{\mathsf{P}}) \; > \; \Delta n^{\textcircled{\raisebox{.5pt}{\sf{\tiny{D}}}}} (\overline{\mathsf{P}})$

By substitution, this implies that if $\mathsf{P}$ is right-handed then $\overline{\mathsf{P}}$ is left-handed, and vice versa. The foregoing definition implies that particles and their conjugate twins always have opposite handedness. We can write

$\delta_{\! \concavediamond} ( \mathsf{P} ) = - \delta_{\! \concavediamond} ( \overline{\mathsf{P}} )$

The handedness is relevant because sometimes rotational phenomena involve many particles in complicated arrangements that are difficult to analyze. But handedness is a way of describing rotation that is broader than chirality and the spin-related characteristics. Handedness can be directly related to additional particles that incrementally add more complexity.

So for simpler descriptions we sparingly add a few more quarks to our particle models. More exactly, we put stereochemical quarks into models of nuclear particles by adding stereoisomers.

So far, our models of nuclear particles have been made exclusively from thermodynamic quarks. But now we develop them further so that they also have stereochemical quarks. Here is an essential rule for choosing from among the many possibilities: We require that a particle’s handedness is the same as its chirality. The prerequisite is stated mathematically by

$\delta_{\! \concavediamond} \! = \delta_{\mathsf{o}}$

This semantic selection-rule guarantees conceptual coordination between analytic regimes. It ensures that the terms dextro and levo have the same meaning when applied to nuclear particles as they do when applied to atoms and molecules.

Thus particles composed from levo-quarks will be levorotatory and left-handed. And particles made from dextro-quarks will be dextrorotatory and right-handed. All of our preferred models for nuclear particles follow this rule.

Sensory interpretation: The characteristic of handedness can be associated with geometric isomerism. We show this later for the molecule diazine. A sensory difference between geometric isomers may be experienced directly by comparing the taste of crushed spearmint leaves with spicy caraway seeds. These two flavours are due to left-handed and right-handed forms of the molecule carvone.

The Helicity

Here is way of describing rotation in an ordinary classroom or laboratory that has a reference-frame F with Cartesian coordinates. It employs the spin angular-momentum vector $\overline{\mathrm{S}}$ and a linear momentum vector ${ {\overline{p}}} .$

If the frame F is grounded, then the linear-momentum can be written as $\overline{p} \! = \! \delta_{\mathfrak{m}} \hspace{1px} p \hspace{2px} \widehat{z}$ where $\delta_{\mathfrak{m}}$ marks the matter-type, $p$ is the norm of $\overline{p}$ and $\widehat{z}$ is the central axis unit-vector. This linear-momentum is expressed relative to the frame of reference. So even though $\overline{\mathrm{S}}$ is an intrinsic particle trait, the description of rotation now becomes frame dependent.

Following custom, we define the helicity $\delta_{h} \hspace{1px} \textsf{,}$ for some rotating particle P, as the sign of the projection of $\overline{\mathrm{S}}$ onto ${ {\overline{p}}} .$ Thus

$\delta_{h} \equiv \begin{cases} \hspace{21px} \textsf{zero} &\textsf{if} \hspace{10px} \overline{\mathrm{S}} \cdot \overline{p} = 0 \\ ( \overline{\mathrm{S}} \cdot \overline{p} ) \, / \, | \overline{\mathrm{S}} \cdot \overline{p} | &\textsf{if} \hspace{10px} \overline{\mathrm{S}} \cdot \overline{p} \ne 0 \rule{0px}{15px} \end{cases}$

If $\delta_{h} \! = \! +1$ then P is said to have clockwise helicity. That is, the angular-momentum vector points mostly in the same direction as the linear-momentum vector. And if $\delta_{h} \! = \! -1$ then P is said to have counterclockwise helicity. But if $\delta_{h} \! = \! 0$ then angular and linear momenta are orthogonal and we say that P is not helical.

Two hands and helical spirals — Two kinds of helicity.

As discussed earlier, the spin angular-momenta of any particle $\mathsf{P}$ and its conjugate twin $\overline{\mathsf{P}}$ are related by $\overline{\mathrm{S}} ( \mathsf{P} ) \! = \! \overline{\mathrm{S}} ( \mathsf{\overline{P}} ) .$ But the linear-momentum vector changes direction with the matter-type so $\overline{p} \hspace{1px} ( \mathsf{P} ) \! = \! - \overline{p} \hspace{1px} ( \mathsf{\overline{P}} ) \hspace{1px} .$ Then by substitution

$\delta_{h} ( \mathsf{P} ) = - \delta_{h} ( \mathsf{\overline{P}} )$

This means that if $\mathsf{P}$ is rotating clockwise, then $\overline{\mathsf{P}}$ is moving counterclockwise, and vice versa. Particles and their conjugate twins always have opposing helicities.

In general, since the linear-momentum is frame dependent, so is the helicity. Sometimes the direction of a momentum vector can be reversed by a change of frame. Then the helicity also changes sign. That is, P’s rotation may be described as being either clockwise or counterclockwise depending on the frame of reference. Please see the discussion of momentum for more detail.

To evaluate a particle’s helicity under some more restrictive conditions, note that the foregoing dot product can be written as

$\overline{\mathrm{S}} \cdot \overline{p} \, = \, \left( \delta_{\mathsf{o}} \| \overline{\mathrm{S}} \| \hspace{1px} \widehat{\rho} \hspace{1px} \right) \cdot \left( \delta_{\mathfrak{m}} \| \overline{p} \| \hspace{1px} \widehat{z} \hspace{1px} \rule{0px}{12px} \right) \, = \, \delta_{\mathsf{o}} \delta_{\mathfrak{m}} \hspace{2px} p \hspace{1px} \| \overline{\mathrm{S}} \| \left( \widehat{\rho} \cdot \widehat{z} \hspace{1px} \right)$

Bearings are marked by $\widehat{\rho}$ for the radial direction vector given by

$\widehat{\rho} \equiv \begin{cases} \hspace{6px} (0,0,0) &\textrm{if} \hspace{5px} \overline{\rho} = (0,0,0) \\ \hspace{7px} \overline{\rho} \, / \, \| \hspace{1px} \overline{\rho} \hspace{1px} \| &\textrm{if} \hspace{5px} \overline{\rho} \ne (0,0,0) \end{cases}$

Here $\overline{\rho}$ is P’s radius vector. And $\rho_{z} \! = \overline{\rho} \cdot \widehat{z}$ is its central axis component. So if $\overline{\rho} \! \ne \! (0,0,0)$ then the dot-product can be written as

$\overline{\mathrm{S}} \cdot \overline{p} \, = \delta_{\mathsf{o}} \delta_{\mathfrak{m}} \hspace{3px} p \hspace{1px} \| \overline{\mathrm{S}} \| \hspace{1px} \dfrac{ \rho_{z} }{ \hspace{1px} \| \hspace{1px} \overline{\rho} \hspace{1px} \| }$

If P is not a photon, then $\delta_{\mathsf{o}} \! = \! \pm 1 .$ And if P is not a graviton, then $\delta_{\mathfrak{m}} \! = \! \pm 1 .$ Then the absolute-value of the dot-product is

$\left| \, \overline{\mathrm{S}} \cdot \overline{p} \, \right| = p \hspace{1px} \| \overline{\mathrm{S}} \| \, \dfrac{ \, | \hspace{1px} \rho_{z} \hspace{1px} | \, }{ \| \hspace{1px} \overline{\rho} \hspace{1px} \| }$

Finally, if momenta are not orthogonal then the helicity is given by a ratio of the last two equations

$\delta_{h} = \delta_{\mathsf{o}} \delta_{\mathfrak{m}} \dfrac{\rho_{z}}{ \, | \, \rho_{z} \hspace{1px} | \, }$

This expression shows the significance of $\rho_{z} \hspace{1px} ,$ the central radius. This central-radius is also important in other ways. It has been considered in detail for understanding how nuclear particles are bound together.

Briefly summarizing, if the central-radius is greater than zero, then the central binding energy is negative. Negative binding-energies imply that bound particles have less energy than their separately unbound constituents. And lower energies are associated with more common, long-lived particles. So for conjugate twins, the most stable twin tends to have a negative binding energy and a positive central-radius. This does not apply to particles that are bound thermoelectrically, but usually the most stable twin is called ordinary-matter. That is, $\delta_{\mathfrak{m}} \! = \! +1 .$

Thus if $\rho_{z} \! \ne \! 0$ then the sign of $\rho_{z}$ will often be the same as the matter-type $\delta_{\mathfrak{m}} .$ This is true for electrons, protons, neutrons and atoms. Then the product of $\delta_{\mathfrak{m}}$ with the sign of $\rho_{z}$ is equal to the number one. And therefore, for many particles that satisfy the foregoing conditions, the helicity becomes simply

$\delta_{h} \! = \delta_{\mathsf{o}}$

That is, the helicity, chirality and handedness will all be coordinated: Particles composed from dextro-quarks will be dextrorotatory with a right-handed, clockwise helicity . Whereas particles made from levo-quarks are levorotatory and left-handed, with a counterclockwise helicity.

This semantic rapport is not always available. Indeed, since the chirality is intrinsic, and the helicity is frame-dependent, they cannot be identical for all particles in all frames. Nonetheless, the pattern is useful for ordinary particles in commonplace frames.

 [+]

References
1	M.B. de Kock, S. Azim, G.H. Kassier and R.J.D. Miller, Determining the radial distribution function of water using electron scattering. Journal of Chemical Physics, Volume 153, Issue 19, 2020.
2	Isaac Newton, Mathematical Principles of Natural Philosophy, page 10. Translated by Andrew Motte and Florian Cajori. University of California Press, 1946.
3	Ernst Mach, The Science of Mechanics, second edition page 231. Translated by Thomas J. McCormack. The Open Court Publishing Company, Chicago 1902.
4	This spin-projection quantum-number for the electron can be written-out in terms of the quark coefficients of the hydrogen atom as $m_{s} = ( n^{\mathsf{u}} + n^{\mathsf{\overline{u}}} - 3 n^{\mathsf{d}} + n^{\mathsf{\overline{d}}} ) / 8 \, .$